\(A=\left(2x\right)^2-2.2x.5+5^2-4x.x+4x.6\)

\(=4x^2-20x+25-4x^2+24x=4x+25\)

\(B=\left(7x-3y\right)^2-\left(7x-3y\right)\left(7x+3y\right)\)

\(=\left(7x-3y\right)\left(7x-3y-7x-3y\right)\)

\(=\left(7x-3y\right)\left(-6y\right)=18y^2-42xy\)

\(C=\left(3-2x\right)^2+\left(3+2x\right)^2\)

\(=9-2.3.2x+4x^2+9+2.3.2x+4x^2\)

\(=18+8x^2\)

\(D=\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+x\right)\left(y-z\right)\)

\(=\left(x-y+z+z-y\right)^2=x^2\)

a, x+15=20-4x

=> x + 4x = 20 - 15

=> 5x = 5

=> x = 5 : 5

=> x = 1

b, 17-x=7-6x

=> -x + 6x = 7 - 17

=> 5x = - 10

=> x = -10 : 5

=> x = -2

c, 4.(x-5)-3.(x+7)= -19

=> 4x - 4.5 - 3.x + 3.7 = -19

=> 4x - 20 - 3x + 21 = -19

=> (4x-3x) + (21-20) = -19

=> 1x + 1 = -19

=> x = -19 - 1

=> x = -18

d, -7.(5-x)-2.(x-10)=15

=> -7x - (-7).5 - 2x - 2.10 = 15

=> -7x - (-35) - 2x - 20 = 15

=> -7x + 35 - 2x - 20 = 15

=> -7x+2x + (35-20) = 15

=> -5x + 15 = 15

=> -5x = 15 - 15

=> -5x = 0

=> x = 0 : (-5)

=> x = 0

e, -5.(2-x)+4.(x-3) = 10.x-15

=> -5x - (-5).2 + 4x - 4.3 = 10x - 15

=> -5x - (-10) + 4x - 12 = 10x - 15

=> -5x + 10 + 4x - 12 = 10x - 15

=> (-5x+4x) - (12-10) = 10x - 15

=> -1x - 2 = 10x - 15

=> -1x - 10x = -15 + 2

=> -11x = -13

=> x = \(\frac{-13}{-11}\)

a.2^3x+1=128

=> 2^{3x + 1} = 2⁷

=> 3x + 1 = 7

=> 3x = 6

=> x = 2

vay_

a.2^3x +1=128

= 2^3x x 2=128

 =128:2=64=2 mũ 6

vậy x=2

b.(7x-11)³=2⁵+5²+200

(7x-11)³=257=6,3579 mũ 3

7x=17,3579

x=2,4797

c.(2x+1)+(2x+2)+(2x+3)+...+(2x+101)=5757

vế trái có 101 số hạng

VT =(2x +1+2x+101).101:2=(4x+102).101:2=5757

(4x+102).101 =5757.2=11514

(4x+102)=11514:101=114

4x=114-102=12

x=12:4=3

vậy x=3

Đề bài mình viết thiếu là CM biểu thức sau không phụ thuộc vào x ( nghĩa là kết quả phải ra số tự nhiên không có x ) 

\(A=\left(2x+1\right)\left(x-1\right)-2x\left(x+2\right)-5\left(-x+3\right)+4\)

\(=2x^2-2x+x-1-2x^2-4x+5x-15+4\)

\(=-12\left(đpcm\right)\)

\(B=\left(4x+3\right)\left(2x-5\right)-\left(8x+1\right)\left(x+3\right)+13\left(3x+1\right)+2\)

\(=8x^2-20x+6x-15-\left(8x^2+24x+x+3\right)+39x+13+2\)

\(=-3\left(đpcm\right)\)

-.-? lớp 6 

Có \(\left(5x-7\right)\left(2x+3\right)-\left(7x+2\right)\left(x-4\right)\)

\(=10x^2+15x-14x-21-7x^2+28x-2x+8\)

\(=\left(10x^2-7x^2\right)+\left(15x-14x+28x-2x\right)+\left(-21+8\right)\)

\(=3x^2+27x-13\)

Thay x = 1/2 vào biểu thức đã rút gọn 

\(\Rightarrow3\left(\frac{1}{2}\right)^2+27.\frac{1}{2}-13=\frac{5}{4}\)

Vậy giá trị biểu thức là 5/4

(5x-7)(2x+3)-(7x+2)(x-4)

=10x²+15x-14x-21-(7x²-7x+2x-8)

=10x²-x-21-7x²+7x-2x+8

=3x²+4x-13

Thay x=1/2 vào BT ta được:

3*(1/2)²+4*1/2-13

=3/4+2-13

=3/4-11

=-41/4

a)  \(3\left(4-2x\right)-2\left(x+3\right)=12-7x\)

\(\Leftrightarrow\)\(12-6x-2x-6=12-7x\)

\(\Leftrightarrow\)\(6-8x=12-7x\)

\(\Leftrightarrow\)\(x=-6\)

Vậy...

b)  \(\left|16+\right|3\left(x-2\right)||-5=20\)

\(\Leftrightarrow\)\(\left|16+\right|3\left(x-2\right)||=25\)(1)

Ta thấy:  \(\left|3\left(x-2\right)\right|\ge0\)\(\Rightarrow\)\(16+\left|3\left(x-2\right)\right|>0\)

nên từ (1)   \(\Rightarrow\)  \(16+\left|3\left(x-2\right)\right|=25\)

                   \(\Leftrightarrow\)\(\left|3\left(x-2\right)\right|=9\)

                   \(\Leftrightarrow\)  \(\orbr{\begin{cases}3\left(x-2\right)=9\\3\left(x-2\right)=-9\end{cases}}\)

                  \(\Leftrightarrow\) \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

Vậy....

c)   \(\left|-5-3^2\right|-||3x+5|-7.2^3|=3^9:3^7\)

\(\Leftrightarrow\)\(14-||3x+5|-56|=9\)

\(\Leftrightarrow\)\(||3x+5|-56|=5\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\left|3x+5\right|-56=5\\\left|3x+5\right|-56=-5\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\left|3x+5\right|=61\\\left|3x+5\right|=51\end{cases}}\)

đến đây bn giải tiếp nhé

2x = (-36)

=> x = (-36) : 2

= > x = (-18)

2x + 25 = - 11

=> 2x   = - 11 - 25

=> 2x   = - 36

=>  x    = - 36 : 2

=>  x    = - 18

2x+25=-11

2x=-11-25

2x=-36

x=-36/2

x=-18