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1: \(\Leftrightarrow\left(x+2\right)\left(x-2\right)+3\left(x+1\right)=3+x^2-x-2\)
\(\Leftrightarrow x^2-x+1=x^2-4+3x+3=x^2+3x-1\)
=>-4x=-2
hay x=1/2
2: \(\Leftrightarrow\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)
\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
\(\Leftrightarrow2x^2+23x+61=2x^2+2x+11\)
=>21x=-50
hay x=-50/21
3: \(\Leftrightarrow6\left(x-8\right)+\left(x+2\right)\left(x-5\right)=-18-\left(x-5\right)\left(x-8\right)\)
\(\Leftrightarrow6x-48+x^2-3x-10+18+x^2-13x+40=0\)
\(\Leftrightarrow2x^2-10x=0\)
=>2x(x-5)=0
=>x=0(nhận) hoặc x=5(loại)
a) 0,25x+1,5=0
=> x = (0 - 1,5) : 0,25 = -1,5 : 0,25 = -6
Vậy x = -6.
b) 6,36−5,3x=0
=> x = (0 + 6,36) : 5,3 = 6,36 : 5,3 =\(\dfrac{6}{5}=1,2\)
Vậy x = 1,2.
c) 43x−56=12
=> x = \(\left(\dfrac{1}{2}+\dfrac{5}{6}\right)\): \(\dfrac{4}{3}\) = \(\dfrac{4}{3}:\dfrac{4}{3}=1\)
Vậy x = 1.
d) −59x+1=23x−10
=> \(\dfrac{-5}{9}x-\dfrac{2}{3}x=\dfrac{-11}{9}x=-10-1=-11\)
=> \(x=-11:\dfrac{-11}{9}=9\)
Vậy x = 9.
a: \(x< -9:\dfrac{3}{2}=-9\cdot\dfrac{2}{3}=-6\)
b: 2/3x>-2
hay x>-2:2/3=-3
c: \(2x>\dfrac{9}{5}-\dfrac{4}{5}=1\)
hay x>1/2
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}>6-4=2\)
hay x>2:3/5=2x5/3=10/3
\(\Leftrightarrow\dfrac{2x}{a^2-a+1}+\dfrac{-4x}{2a^2-2a+2a^2}+\dfrac{2ax}{1+a^3}< \dfrac{1}{2a+2}-\dfrac{1}{2a^2-2a+2}+\dfrac{a}{1+a^3}\)
\(\Leftrightarrow\left(\dfrac{2}{a^2-a+1}-\dfrac{4}{2a^2-2a+2}+\dfrac{2a}{1+a^3}\right).x< \left(\dfrac{1}{2a+2}-\dfrac{1}{2a^2-2a+2}+\dfrac{a}{1+a^3}\right)\)
\(\Leftrightarrow\left(\dfrac{2a}{1+a^3}\right)x< \dfrac{\left(a^2-a+1\right)-\left(a+1\right)+2a}{2.\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{a^2}{1+a^3}\)
\(\Leftrightarrow\left(\dfrac{2a}{1+a^3}\right)x< \dfrac{a^2}{2.\left(1+a^3\right)}\)
\(a=0\Rightarrow vo...N_o\)
\(\left\{{}\begin{matrix}\dfrac{2a}{a^3+1}>0\Leftrightarrow\left[{}\begin{matrix}a< -1\\a>0\end{matrix}\right.\\x< \dfrac{a^2}{2\left(a^3+1\right)}:\dfrac{2a}{\left(a^3+1\right)}=\dfrac{a}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2a}{a^3+1}< 0\Rightarrow-1< a< 0\\x>\dfrac{a}{2}\end{matrix}\right.\)
ĐKXĐ: x\(\ne3,x\ne-3\)
\(\Rightarrow\left(x-a\right)\left(a-3\right)+\left(x+3\right)\left(a+3\right)=-6a\)
\(\Leftrightarrow xa-3x-a^2+3a+ax+3x+3a+3=-6a\)
\(\Leftrightarrow2ax-a^2+12a+3=0\) \(\Leftrightarrow2ax=a^2-12a-3\Leftrightarrow x=\dfrac{a^2}{2}-6a-\dfrac{3}{2}\)(TM)
Vậy...
bạn lm sai rồix-3 chứ ko phải x+3 từ dòng thứ 2