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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: ...
\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)
\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)
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) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)