Sửa đề: \(P=\dfrac{2}{2x+3}+\dfrac{3}{2x+1}-\dfrac{6x+5}{\left(2x+3\right)\left(2x+1\right)}\)

a: ĐKXĐ: \(x\notin\left\{-\dfrac{3}{2};-\dfrac{1}{2}\right\}\)

b: \(A=\dfrac{4x+2+6x+9-6x-5}{\left(2x+3\right)\left(2x+1\right)}=\dfrac{4x+6}{\left(2x+3\right)\left(2x+1\right)}=\dfrac{2}{2x+1}\)

c: Để P=-1 thì 2x+1=-2

=>2x=-3

hay x=-3/2(loại)

a.

P được xác định khi \(\left[{}\begin{matrix}2x+3=0\\2x-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)

vậy ĐKXĐ là: \(x\ne\pm\dfrac{3}{2}\)

b.

\(P=\dfrac{2}{2x+3}+\dfrac{3}{2x-3}-\dfrac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\\ P=\dfrac{2\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\dfrac{3\left(2x+3\right)}{\left(2x+3\right)\left(2x-3\right)}-\dfrac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)

\(P=\dfrac{2\left(2x-3\right)+3\left(2x+3\right)-6x-5}{\left(2x+3\right)\left(2x-3\right)}\\ P=\dfrac{4x-6+6x+9-6x-5}{\left(2x+3\right)\left(2x-3\right)}=\dfrac{4x-2}{\left(2x+3\right)\left(2x-3\right)}\)

c.

theo đề bài, ta có:

\(\dfrac{4x-2}{\left(2x+3\right)\left(2x-3\right)}=4\\ \Leftrightarrow4x-2=4\left(2x+3\right)\left(2x-3\right)\)

\(\Leftrightarrow4x-2=4\left(4x^2-6x+6x-9\right)\\ \Leftrightarrow2x-1=8x^2-18\)

\(\Leftrightarrow8x^2-2x-17=0\\ \Leftrightarrow x^2-\dfrac{1}{4}x=\dfrac{17}{8}\)

\(\Leftrightarrow x^2-2.\dfrac{1}{8}+\dfrac{1}{64}=\dfrac{17}{8}+\dfrac{1}{64}\\ \Leftrightarrow\left(x-\dfrac{1}{8}\right)^2=\dfrac{137}{64}\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{8}=\dfrac{\sqrt{137}}{8}\\x-\dfrac{1}{8}=-\dfrac{\sqrt{137}}{8}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{137}+1}{8}\\x=\dfrac{1-\sqrt{137}}{8}\end{matrix}\right.\)

vậy P=4 khi \(x=\dfrac{\sqrt{137}+1}{8}\) và \(x=\dfrac{1-\sqrt{137}}{8}\)

\(P=\dfrac{2}{2x+3}+\dfrac{3}{2x-3}-\dfrac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)

ĐKXĐ \(2x+3\ne0\) và \(2x-3\ne0\)

Suy ra \(x\ne\dfrac{-3}{2}\) và \(x\ne\dfrac{3}{2}\)

MC: (2x+3)(2x-3)

\(\dfrac{2.\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\dfrac{3.\left(2x+3\right)}{\left(2x+3\right)\left(2x-3\right)}-\dfrac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)

\(=\dfrac{4x-6}{\left(2x+3\right)\left(2x-3\right)}+\dfrac{6x+9}{\left(2x+3\right)\left(2x-3\right)}\dfrac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)

\(=\dfrac{4x-6}{\left(2x+3\right)\left(2x-3\right)}\)

a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b) Ta có: \(A=\left(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+2}{2x+2}\right)\cdot\dfrac{2x^2-2}{5}\)

\(=\left(\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\dfrac{6}{2\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+2\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\right)\cdot\dfrac{2x^2-2}{5}\)

\(=\left(\dfrac{x^2+2x+1+6-\left(x^2-x+2x-2\right)}{2\left(x+1\right)\left(x-1\right)}\right)\cdot\dfrac{2x^2-2}{5}\)

\(=\dfrac{x^2+2x+7-x^2-x+2}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{x+9}{5}\)

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: ...

Đề sai rồi bạn

\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)

\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

\(=\dfrac{\left(2x-3y\right)^3}{\left(2x-3y\right)^2}=2x-3y\)

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3

a:

Sửa đè: \(B=\left(2x+1+\dfrac{1}{2x-1}\right):\left(\dfrac{2x^2-6x}{x-3}-\dfrac{4x^2}{2x-1}\right)\)

 \(B=\dfrac{4x^2-1+1}{2x-1}:\left(2x-\dfrac{4x^2}{2x-1}\right)\)

\(=\dfrac{4x^2}{2x-1}:\dfrac{4x^2-2x-4x^2}{2x-1}\)

\(=\dfrac{4x^2}{-2x}=-2x\)

b: |x-2|=1

=>x-2=1 hoặc x-2=-1

=>x=1(nhận) hoặc x=3(loại)

Khi x=1 thì A=-2*1=-2

a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)

c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)

d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)

e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)