Đề thi tuyển sinh vào lớp 10 môn Toán chuyên năm 2022-2023 sở GD&ĐT Hưng Yên có đáp án

a) $$A=\left(\frac{x+2\sqrt{x}}{x+\sqrt{x}-2}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}$$                                 ĐK: $x\ge 0,x\ne 1$

                = $$\left(\frac{\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{2}{\sqrt{x}(\sqrt{x}-)}\right):\frac{1}{\sqrt{x}-1}$$

                = $$\left(\frac{x+2}{(\sqrt{x}-1)\sqrt{x}}.\frac{\sqrt{x}-1}{1}\right)$$

                = $$\frac{x+2}{\sqrt{x}}$$

b) $A=3\iff \frac{x+2}{\sqrt{x}}=3$

                $\iff x+2=3\sqrt{x}$

                $\iff x-3\sqrt{x}+2=0$

                $$\iff [\begin{array}{l}\sqrt{x}=1\\ \sqrt{x}=2\end{array}\iff [\begin{array}{l}x=1(l)\\ x=4(n)\end{array}$$

        Vậy A = 3 khi x = 4.

1) Hoành độ giao điểm của (P) và (d)

$$x^2=(m+1)x-m+5$$

$$\iff x^2-(m+1)x+m-5=0$$(*)

Ta có $$\Delta =(m+1{)}^2-4(m-5)$$

$$=m^2-2m+21$$

$$=(m-1{)}^2+20>0$$

Nên (d) luôn cắt (P) tại hai điểm phân biệt A và B

Theo hệ thức vi-et $\{\begin{array}{l}{x}_1+x_2=m+1\\ x_1.x_2=m-5\end{array}$

        (*) $\iff x^2-x-5=m(x-1)$

Xét $x=1$ không phải là nghiệm của phương trình

$\Rightarrow x-\frac{5}{x-1}=m$  (1)

Vì $x_1;x_2\in Z$nên $m+1$ và $m-5$ là các số nguyên do đó $m$cũng là số nguyên

Từ (1) ta có 

  $m\in Z$ khi $\left(x-\frac{5}{x-1}\right)\in Z$ $\Rightarrow \{\begin{array}{l}x-1\in Z\\ 5⋮\left(x-1\right)\end{array}$

Suy ra $m=-3;m=5$

$[\begin{array}{l}x-1=-5\iff x=-4\Rightarrow m=-3\\ x-1=-1\iff x=0\Rightarrow m=5\\ x-1=5\iff x=6\Rightarrow m=5\\ x-1=1\iff x=2\Rightarrow m=-3\end{array}$

Vậy $m=-3;m=5$ thỏa yêu cầu bài toán

2) $$x^4-2x^3+x^2-16y^2+12x-16y+4=0$$

            $$\iff x^4+x^3-3x^3-3x^2+4x^2+4x+8x+8=16y^2+16y+4$$

            $$\iff (x+1)x^3+3x^2(x+1)+4x(x+1)+8(x+1)=16y^2+16y+4$$

            $\iff (x+1)(x^3-3x^2+4x+8)=(4y+2{)}^2$

            $\iff (x+1{)}^2(x^2-4x+8)=(4y+2{)}^2$

Vì $y\in z\Rightarrow 4y+2\ne 0\Rightarrow x\ne -1$

Vì $x,y\in z$ nên $(x+1{)}^2$ và $(4y+2{)}^2$ là số chính phương khác 0 nên $(x^2-4x+8)$ cũng là số chính phương

Đặt $x^2-4x+8=m$ $(m\in N^{\ast })$

            $\iff (x-2{)}^2+4=m^2$

            $\iff (x-2{)}^2-m^2=-4$

            $\iff (x-2-m)(x-2+m)=-4^{(\ast )}$

Do $x-2-m<x-2+m$

 $x=-2\Rightarrow (4y+2{)}^2=4$ $$\iff [\begin{array}{l}4y+2=2\\ 4y+2=-2\end{array}\iff [\begin{array}{l}4y=0\\ 4y=-4\end{array}\iff [\begin{array}{l}y=0\\ y=-1\end{array}$$

Vậy nghiệm nguyên thỏa ycbt là: (– 2; 0); (– 2; – 1).

1) $$\sqrt{\frac{3x-2}{x-1}}-\sqrt{\frac{3-x}{x-1}}=1$$

ĐK: $1<x\le 3$

$\iff \sqrt{3x-2}-\sqrt{3-x}=\sqrt{x-1}$

$\iff \sqrt{3x-2}=\sqrt{x-1}+\sqrt{3-x}$

$\iff 3x-2=x-1+3-x+2\sqrt{(x-1)(3-x)}$

$\iff 3x-4=2\sqrt{(x-1)(3-x)}$ (*)

(*) có điều kiện: $3x-4\ge 0\iff x\ge \frac{4}{3}$

(*) $\iff 9x^2-24x+16=4(x-1)(3-x)$

$\iff 9x^2-24x+16=-4x^2+16x-12$

$\begin{array}{l}\iff 13x^2-40x+28=0\\ \iff [\begin{array}{l}x=2(n)\\ x=\frac{14}{13}(l)\end{array}\end{array}$

Vậy nghiệm của phương trình: $x=2$.

2)

$\{\begin{array}{l}{x}^3+y^3+xy=2x+4y-1\\ xy+x+2y=1\end{array}$$\iff \{\begin{array}{l}{x}^3+y^3+xy=2(1-xy)-1\\ x+2y=1-xy\end{array}$ $\iff \{\begin{array}{l}{x}^3+y^3+3xy=1(1)\\ x+2y=1-xy(2)\end{array}$

$\left(1\right)\iff (x+y{)}^3-3x^{2}y-3x{y}^2+3xy-1=0$

$\begin{array}{l}\iff {\left(x+y\right)}^3-1^3-3xy(x+y-1)=0\\ \iff \left(x+y-1\right)\left[{\left(x+y\right)}^2+x+y+1-3xy\right]\\ \iff \left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)=0\end{array}$

$\iff [\begin{array}{l}x+y-1=0\\ x^2+y^2-xy+x+y+1=0\end{array}$

Với $x+y-1=0\iff x=1-y$ thay vào (2) ta được:

$\left(2\right)\iff$$1-y+2y=1-(1-y)y$$\iff y\left(y-2\right)=0$

$\iff [\begin{array}{l}y=0\Rightarrow x=1\\ y=2\Rightarrow x=-1\end{array}$

Với : $x^2+y^2-xy+x+y+1=0\iff 2x^2+2y^2-2xy+2x+2y+2=0$

$\begin{array}{l}\iff \left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2+2y+1\right)=0\\ \iff {\left(x-y\right)}^2+{\left(x+1\right)}^2+{\left(y+1\right)}^2=0\end{array}$

$\iff \{\begin{array}{l}x-y=0\\ x+1=0\\ y+1=0\end{array}\iff x=y=-1$

Vậy nghiệm của hệ phương trình là: $(1;0),(2;-1),(-1;-1)$.

Ta có: $4xy+2yz+3xz=24\iff \frac{xy}{6}=\frac{yz}{12}=\frac{xz}{8}=1$$\iff \frac{x}{2}.\frac{y}{3}+\frac{y}{3}.\frac{z}{4}+\frac{x}{2}.\frac{z}{4}=1$         

Đặt $\frac{x}{2}=a>0;\frac{y}{3}=b>0;\frac{z}{4}=c>0\Rightarrow ab+bc+ac=1$

$\begin{array}{l}P=\frac{4a}{\sqrt{4a^2+4}}+\frac{3b}{\sqrt{9b^2+9}}+\frac{4c}{\sqrt{16c^2+16}}\\ =\frac{2a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\end{array}$

$=\frac{2a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ac}}+\frac{c}{\sqrt{c^2+ab+bc+ac}}$

$=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}$

$=\sqrt{\frac{2a}{a+b}.\frac{2a}{a+c}}+\sqrt{\frac{2b}{a+b}.\frac{b}{2\left(b+c\right)}}+\sqrt{\frac{c}{2\left(b+c\right)}.\frac{2c}{a+c}}$

Ta có :

$\begin{array}{l}\frac{2a}{a+b}+\frac{2a}{a+c}\ge 2\sqrt{\frac{2a}{a+b}.\frac{2a}{a+c}}\\ \frac{2b}{a+b}+\frac{b}{2\left(b+c\right)}\ge 2\sqrt{\frac{2b}{a+b}.\frac{b}{2\left(b+c\right)}}\\ \frac{c}{2\left(b+c\right)}+\frac{2c}{a+c}\ge 2\sqrt{\frac{c}{2\left(b+c\right)}.\frac{2c}{a+c}}\end{array}$

$p\le \frac{1}{2}\left(\frac{2a}{a+b}+\frac{2a}{a+c}+\frac{2b}{a+b}+\frac{b}{2\left(b+c\right)}+\frac{c}{2\left(b+c\right)}+\frac{2c}{a+c}\right)$

  $\begin{array}{l}\iff P\le \frac{1}{2}\left(\frac{2\left(a+b\right)}{a+b}+\frac{2\left(a+c\right)}{a+c}+\frac{b+c}{2(b+c)}\right)\\ \iff P\le \frac{1}{2}\left(2+2+\frac{1}{2}\right)\\ \iff P\le \frac{9}{4}\end{array}$

Dấu  xảy ra khi và chỉ khi $\{\begin{array}{l}\frac{2a}{a+b}=\frac{2a}{a+c}\\ \frac{2b}{a+b}=\frac{b}{2\left(b+c\right)}\\ \frac{c}{2\left(b+c\right)}=\frac{2c}{a+c}\end{array}\iff \{\begin{array}{l}b=c\\ a+b=8b\\ a+c=8c\end{array}\iff \{\begin{array}{l}b=c\\ a=7b\\ a=7c\end{array}$

$ab+bc+ac=1\iff 7b^2+b^2+7b^2=1\iff b^2=\frac{1}{15}\iff b=\frac{1}{\sqrt{15}}$

$\{\begin{array}{l}b=\frac{1}{\sqrt{15}}\Rightarrow y=\frac{3}{\sqrt{15}}\\ c=\frac{1}{\sqrt{15}}\Rightarrow z=\frac{4}{\sqrt{15}}\\ a=\frac{7}{\sqrt{15}}\Rightarrow x=\frac{14}{\sqrt{15}}\end{array}$.