1: A=4x^2+12x+9-4x^2+4x-1-6x=10x+8

Khi x=201 thì A=10*201+8=2018

2: B=4x^2+20x+25-4x^2+12=20x+37

Khi x=1/20 thì B=1+37=38

1, \(A=\left(2x+3\right)^2-\left(2x-1\right)^2-6x\)

\(A=\left[\left(2x+3\right)+\left(2x-1\right)\right]\left[\left(2x+3\right)-\left(2x-1\right)\right]-6x\)

\(A=\left(2x+3+2x-1\right)\left(2x+3-2x+1\right)-6x\)

\(A=4\left(4x+2\right)-6x\)

\(A=16x+8-6x\)

\(A=10x+8\)

Thay \(x=201\) vào A ta có:

\(A=10\cdot201+8=2010+8=2018\)

Vậy: ....

2, \(B=\left(2x+5\right)^2-4\left(x+3\right)\left(x-3\right)\)

\(B=\left(2x+5\right)^2-4\left(x^2-9\right)\)

\(B=4x^2+20x+25-4x^2+36\)

\(B=20x+61\)

Thay \(x=\dfrac{1}{20}\) vào B ta có:

\(B=20\cdot\dfrac{1}{20}+61=1+61=62\)

Vậy: ...

\(M=\left(7-2x\right)\left(4x^2+14x+49\right)-\left(64-8x^3\right)\)

\(M=\left(7-2x\right)\left[\left(2x\right)^2+2x\cdot7+7^2\right]-\left(64-8x^3\right)\)

\(M=\left[7^3-\left(2x\right)^3\right]-\left(64-8x^3\right)\)

\(M=343-8x^3-64+8x^3\)

\(M=279\)

Vậy M có giá trị 279 với mọi x

\(P=\left(2x-1\right)\left(4x^2-2x+1\right)-\left(1-2x\right)\left(1+2x+4x^2\right)\)

\(P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3\)

\(P=16x^3-8x^2+4x-2\)

Thay \(x=10\) vào P ta có:

\(P=16\cdot10^3-8\cdot10^2+4\cdot10-2=15238\)

Vậy P có giá trị 15238 tại x=10

a: M=343-8x^3-64+8x^3=279

b: P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3

=16x^3-8x^2+4x-2

=16*10^3-8*10^2+4*10-2=15238

thiếu đề : \(\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}.\)

Bài 2 :

a, Để \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4^2-4}{5}\)

\(\Rightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}}\Rightarrow\orbr{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)

b,\(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4x^2-4}{5}\)

\(B=\left[\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x+1\right)\left(x-1\right)}-\frac{x+3}{2\left(x+1\right)}\right].\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\left[\frac{x^2+2x+1}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\left[\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\frac{4}{2\left(x-1\right)\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\frac{8}{5}\)

=> giá trị của B ko phụ thuộc vào biến x

bài 1

=\(^{\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x+1\right)^2}\)

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

=\(\left(4x\right)^2\)

=\(16x^2\)

Tại x=100 thay vào biểu thức trên ta có:

16*100^2=1600000

\(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)=\left[\frac{x+1}{2.\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2.\left(x+1\right)}\right]\)

\(\Rightarrow\hept{\begin{cases}x\ne1\\x\ne\pm1\\x\ne-1\end{cases}\Rightarrow x\pm1}\)

Vậy để B xác định => x=+-1

Giups mik vs

chịu

Đề sai rồi bạn

a)  M = ( 2x + 3)(2x - 3) - 2(x + 5)² - 2(x - 1)(x + 2) 

   = 4x² - 9 - 2(x² + 10x + 25) - 2(x² + x - 2)

   = 4x² - 9 - 2x² - 20x - 50 - 2x² - 2x + 4

   = -22x - 55 =  -11(2x + 5)

b) M = -11(2x + 5) = - 11(2.\(\frac{-7}{3}\)+ 5) = \(\frac{-11}{3}\)

b)  M = -11(2x + 5) = 0

\(\Rightarrow\)2x + 5 = 0

\(\Rightarrow\)x = \(\frac{-5}{2}\)

Ta có: M = (2x+3)(2x-3) - 2(x+5)² - 2(x-1)(x+2) \(=\left(2x\right)^2-3^2-2\left(x^2+10x+25\right)-\) \(2\left(x^2+x-2\right)\)

\(=4x^2-9-2x^2-20x-50-2x^2-2x+4\) =\(\left(4x^2-2x^2-2x^2\right)-\left(20x+2x\right)-\left(50+9-4\right)\) \(=-22x-55\)

b, Với x = \(-2\frac{1}{3}=\frac{-7}{3}\)

\(\Rightarrow M=-22.\frac{-7}{3}-55=\frac{154}{3}-55=\frac{-11}{3}\)

c, Để M = 0 => -22x - 55 = 0 \(\Rightarrow-22x=55\Rightarrow x=\frac{-55}{22}=\frac{-5}{2}\)

Vậy \(x=\frac{-5}{2}\)