Bài 1: 

\(101\cdot125+101\cdot25-101\cdot50\)

\(=101\cdot\left(125+25-50\right)\)

\(=101\cdot100\)

\(=10100\)

Bài 2: 

\(76\cdot115+56\cdot24+59\cdot24\)
\(=76\cdot115+24\cdot\left(56+59\right)\)

\(=76\cdot115+24\cdot115\)

\(=115\cdot\left(76+24\right)\)

\(=115\cdot100\)

\(=11500\)

5:

a: =>15x=165

=>x=11

b: =>x(x+1)/2=55

=>x^2+x=110

=>x=10

4: =>7x=56

=>x=8

Bài 1:

101•125+101•25+101•50

= 101•(125+25-50)

=101•100

=10100

Bài 2:

76•115+56•24+59•24

= 76•115+24•(56+59)

= 76•115+24•115

= 115•(76+24)

= 115•100

= 11500

x-y=5

nên x=y+5

Ta có: xy=66

=>y(y+5)=66

\(\Leftrightarrow y^2+5y-66=0\)

=>(y+11)(y-6)=0

=>y=-11 hoặc y=6

=>x=-6 hoặc x=11

Bài 1:

(X + 1)(3y - 2) = -55

= -55 . 1 = -1.55 = -11.5 = -5.11

Liệt kê thành bảng

Bài 2:

a) |2x-  5| = 13

2x - 5 = 13 => x = 9

2x - 5 = -13 => x = -4

b) |7x +3| = 66

7x + 3 = 66 => x = 9

7x + 3 = -66 => x = -69/7

Bài 1

\(d,66\cdot25+5\cdot66+66\cdot14+33\cdot66\)

\(=66\left(25+5+14+33\right)\)

\(=66\cdot77\)

\(=5082\)

Hình như chép sai đề vì không thể nào tính nhanh đc

Bài 2

\(7+\left(63:x-11\right)\cdot23=237\)

\(\Leftrightarrow\left(63:x-11\right)\cdot23=230\)

\(\Leftrightarrow63:X-11=10\)

\(\Leftrightarrow63:x=21\)

\(\Leftrightarrow x=3\)

Bài 1 :

\(d,66.25+5.66+66.14+33.66\)

\(=66.\left(25+5+14+33\right)\)

\(=66.77\)

\(=5082\)

Bài 2 :

\(7+\left(63:x-11\right).23=237\)

\(\Leftrightarrow\left(63:x-11\right).23=230\)

\(\Leftrightarrow63:x-11=10\)

\(\Leftrightarrow63:x=21\)

\(\Leftrightarrow x=3\)

Vậy x = 3 .

Học tốt nhé

66.25+5.66+66.14+33.66

=66.(25+5+14+33)

=66.77

=5082

7+(63:x-11).23=237

(63:x-11).23=237-7

(63:x-11).23=230

(63:x-11)=230:23

(63:x-11)=10

63:x=10+11

63:x=21

x=63:21

x=3

vậy x=3

`@` `\text {Ans}`

`\downarrow`

`a)`

`(2x - 1)^2 + 1 = 26`

`\Rightarrow (2x - 1)^2 = 26 - 1`

`\Rightarrow (2x - 1)^2 = 25`

`\Rightarrow (2x - 1)^2 = (+-5)^2`

`\Rightarrow`\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=6\div2\\x=-4\div2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-2;3\right\}\)

`b)`

`(2x - 4)^3 + 2 = 66`

`\Rightarrow (2x - 4)^3 = 66 - 2`

`\Rightarrow (2x - 4)^3 = 64`

`\Rightarrow (2x - 4)^3 = 4^3`

`\Rightarrow 2x - 4 = 4`

`\Rightarrow 2x = 8`

`\Rightarrow x = 8 \div 2`

`\Rightarrow x = 3`

Vậy, `x = 3`

`c)`

\(7^{x+2}+5\cdot7^{x+1}+15=603\)

`\Rightarrow 7^x . 7^2 + 5 . 7^x . 7 = 603 - 15`

`\Rightarrow 7^x . 7^2 + 35 . 7^x = 588`

`\Rightarrow 7^x . (7^2 + 35) = 588`

`\Rightarrow 7^x . 84 = 588`

`\Rightarrow 7^x = 588 \div 84`

`\Rightarrow 7^x = 7`

`\Rightarrow 7^x = 7^1`

`\Rightarrow x = 1`

Vậy, `x = 1.`

\(#48Cd\)

\(a)\left(2x-1\right)^2+1=26\)

\(\left(2x-1\right)^2=25\)

\(TH1:2x-1=5\)

\(2x=6\)

\(x=3\)

\(TH2:2x-1=-5\)

\(2x=-4\)

\(x=-2\)

Vậy........

\(b)\left(2x-4\right)^3+2=66\)

\(\left(2x-4\right)^3=64=4^3\)

\(2x-4=4\)

\(2x=8\)

\(x=4\)

\(c)7^x+2+5.7^x+1+15=603\)

\(7^x\left(1+5\right)=603-15-1-2\)

\(7^x.6=585\)

Bạn xem lại phần này nhé . x tìm ra không được chẵn lắm á cậu.

Bạn xem lại câu a

a,(2x-1)²+1=26

=>(2x-1)²=26-1=25=5²

=>2x-1=5 và -5

=>x=(5+1):2=3 và x=(-5+1):2=-2

b,(2x-4)³+2=66

=>(2x-4)³=66-2=64=4³

=>2x-4=4

=>x=(4+4):2=4

c,7^x+2+5\(\cdot\)7^x+1+15=603

=>7^x\(\cdot\)7²+5\(\cdot\)7^x\(\cdot\)7=603-15=588

=>7^x(7²+5\(\cdot\)7)=588

=>7^x\(\cdot\)84=588

=>7^x=588:84=7

=>x=1

Bài 4:

a: xy=-2

=>\(x\cdot y=1\cdot\left(-2\right)=\left(-2\right)\cdot1=\left(-1\right)\cdot2=2\cdot\left(-1\right)\)

=>\(\left(x,y\right)\in\left\{\left(1;-2\right);\left(-2;1\right);\left(-1;2\right);\left(2;-1\right)\right\}\)

b: \(\left(x-1\right)\left(y+2\right)=-3\)

=>\(\left(x-1\right)\cdot\left(y+2\right)=1\cdot\left(-3\right)=\left(-3\right)\cdot1=-1\cdot3=3\cdot\left(-1\right)\)

=>\(\left(x-1;y+2\right)\in\left\{\left(1;-3\right);\left(-3;1\right);\left(-1;3\right);\left(3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;-5\right);\left(-2;-1\right);\left(0;1\right);\left(4;-3\right)\right\}\)

Bài 3:

a: \(x\left(x+9\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x+9=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)

b: \(\left(x-5\right)^2=9\)

=>\(\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3+5=8\\x=-3+5=2\end{matrix}\right.\)

c: \(\left(7-x\right)^2=-64\)

mà \(\left(7-x\right)^2>=0\forall x\)

nên \(x\in\varnothing\)

Bài 2:

a: \(\left(-31\right)\cdot x=-93\)

=>\(31\cdot x=93\)

=>\(x=\dfrac{93}{31}=3\)

b: \(\left(-4\right)\cdot x=-20\)

=>\(4\cdot x=20\)

=>\(x=\dfrac{20}{4}=5\)

c: \(5x+1=-4\)

=>\(5x=-4-1=-5\)

=>\(x=-\dfrac{5}{5}=-1\)

d: \(-12x+1=-4\)

=>\(-12x=-4-1=-5\)

=>\(12x=5\)

=>\(x=\dfrac{5}{12}\)