Đề thi cuối kì 1 Toán 6 Cánh diều (2023-2024) – Đề 5 (có đáp án)

$72+\left(-190\right)+\left(-72\right)+290$

$=72-192-72+290$

$=\left(72-72\right)+\left(290-190\right)$

$47\cdot 13+47\cdot 12-25\cdot 147$ = (47.13+47.12) -25.147

= 47.(13 +12) -25.147 = 47.25 – 25 .147 = 25.( 47 – 147)

$$\left(-25\right)\cdot 7\cdot \left(-4\right)\cdot 3$$

$132+2\left(x-4\right)=46$

$2\left(x-4\right)=46-132$

$$\begin{array}{l}x-4=-86:2\\ x=-43+4\\ <\mathrm{span}></span>x=-39\end{array}$$

$$19-2x=7^5:7^3$$

$$19-2x=7^{5-3}$$

$$19-2x=7^2$$

$$19-2x=49$$

        $$2x=19-49$$

       $$2x=-30$$

        $$x=-15$$

$$18⋮x;42⋮x$$ và $$-3\le x<2.$$

Ta có: $$18⋮x;42⋮x$$ nên $$x\in$$ƯC$$\left(18,42\right)$$

Ta có $$18=2\cdot 3^2;42=2\cdot 3\cdot 7$$.

ƯCLN$$\left(18,42\right)=2\cdot 3=6$$

ƯC$$\left(18,42\right)$$ = Ư$$\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}$$

Gọi số học sinh của trường đó là a học sinh $\left(a\in \mathbb{N}\ast \right)$.\(\)

Ta có $$a⋮12;a⋮18;a⋮21$$ và$$500\le a\le 600$$                                                

Vì $$a⋮12;a⋮18;a⋮21$$ nên $$a\in BC\left(12,18,21\right)$$.

Có $$12=2^2\cdot 3;18=2\cdot 3^2;21=3\cdot 7$$

Do đó $$BCNN\left(12,18,21\right)=2^2\cdot 3^2\cdot 7=252$$.      

$$BC\left(12,18,21\right)=B\left(252\right)=\left\{0;252;504;756;...\right\}$$.           

Vì $$a\in BC\left(12,18,21\right)$$ và $$500\le a\le 600$$nên $$a=504$$.

Có $$3n+7=3n-3+10=3\left(n-1\right)+10$$

Do $$3\left(n-1\right)⋮\left(n-1\right)$$ nên $$10⋮\left(n-1\right)$$.

Suy ra $$n-1\in$$ Ư$$\left(10\right)=\left\{-10;-5;-2;-1;1;2;5;10\right\}$$

Gọi ƯCLN$\left(7n+10,5n+7\right)=d$.

$7n+10⋮d;5n+7⋮d$

$$35n+50⋮d;35n+49⋮d$$

Suy ra $$\left[\left(35n+50\right)-\left(35n+49\right)\right]⋮d$$

Do đó $$1⋮d$$ hay $$d=1$$.