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Bài 1:
a: =>2x-9=10/91
=>2x=829/91
hay x=829/182
b: =>2x=-7
hay x=-7/2
c: =>-3x=-12
hay x=4
(x-1)(2x^2-8)=0
\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)
\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)
3x^2-8x+5=0
áp dụng công thức bậc 2 ta có:
\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)
\(\Rightarrow x=\dfrac{5}{3};x=1\)
(7x-1).2x-7x+1=0
\(\Leftrightarrow\left(7x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)
a: \(\Leftrightarrow\dfrac{y+5}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)
\(\Leftrightarrow2\left(y+5\right)^2-\left(y-5\right)^2=y^2+25y\)
=>\(2y^2+20y+50-y^2+10y-25=y^2+25y\)
=>30y+25=25y
=>5y=-25
=>y=-5(loại)
b: \(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
=>x^2+x+x^2-3x-4x=0
=>2x^2-6x=0
=>2x(x-3)=0
=>x=0(nhận) hoặc x=3(loại)
c: =>x^2-9-6(2x+7)=-13(x+3)
=>x^2-9-12x-42+13x+39=0
=>x^2+x-6=0
=>(x+3)(x-2)=0
=>x=2(nhận) hoặc x=-3(loại)
Ta có : \(\frac{x-1}{-1}+\frac{1}{x+1}=\frac{2x-1}{x^2+x}\)
\(\Rightarrow\) \(-x+1=\frac{2x-1}{x\left(x+1\right)}-\frac{1}{x+1}\) \(=\frac{1}{x+1}.\left(\frac{2x-1-x}{x}\right)=\frac{1}{x+1}.\frac{x-1}{x}\)
\(\Rightarrow\frac{1}{x+1}.\frac{x-1}{x}+x-1=0\)
\(\Rightarrow\frac{x-1}{x\left(x+1\right)}+\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{x\left(x+1\right)}+1\right)=0\)
\(\Rightarrow x-1=0\)hoặc \(\frac{1}{x\left(x+1\right)}+1=0\)
\(\Rightarrow x=1\) hoặc \(\frac{1}{x\left(x+1\right)}=-1\)
\(\Rightarrow x=1\)hoặc \(x^2+x=-1\)
\(\Rightarrow x=1\)hoặc \(x^2+1+x=0\)
\(\Rightarrow x=1\)hoặc \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\)
\(\Rightarrow x=1\)hoặc \(\left(x+\frac{1}{2}\right)^2=-\frac{3}{4}\)(vô lí)
Vậy : x = 1 thì thỏa mãn điều kiện bài toán.
a: =>3(3x-7)+2(x+1)=-96
=>9x-21+2x+2=-96
=>11x-19=-96
=>11x=-96+19=-75
=>x=-75/11
b: \(x-\dfrac{x+1}{3}=\dfrac{2x+1}{5}\)
=>15x-5(x+1)=3(2x+1)
=>15x-5x-5=6x+3
=>10x-5=6x+3
=>4x=8
=>x=2
a)
\(\dfrac{3x-7}{2}+\dfrac{x+1}{3}=-16\)
\(< =>9x-21+2x+2=-96\)
\(< =>9x+2x=-96+21-2\\ < =>11x=-77\\ < =>x=-7\)
b)
\(\dfrac{x-x+1}{3}=\dfrac{2x+1}{5}\\ < =>5=6x+3\\ < =>6x=5-3\\ < =>6x=2\\ < =>x=\dfrac{1}{3}\)
\(\frac{x-2}{5}+7=\frac{x+1}{3}-\frac{8-x}{10}\)
\(\Leftrightarrow\frac{x-2+35}{5}=\frac{10x+10}{30}-\frac{24-3x}{30}\)
\(\Leftrightarrow\frac{x+33}{5}=\frac{13x-14}{30}\Leftrightarrow\frac{6x+198}{30}=\frac{13x-14}{30}\)
Khử mẫu : \(6x+198=13x-14\)
\(\Leftrightarrow-7x=-14-198=212\Leftrightarrow x=-\frac{212}{7}\)
\(\frac{x-2}{5}+7=\frac{1+x}{3}-\frac{8-x}{10}\)
\(\Leftrightarrow\frac{x-2}{5}+\frac{35}{5}=\frac{10\left(1+x\right)}{30}-\frac{3\left(8-x\right)}{30}\)
\(\Leftrightarrow\frac{x-2+35}{5}=\frac{10+10x-24+3x}{30}\)
\(\Leftrightarrow\frac{x+33}{5}=\frac{13x-14}{30}\)
\(\Leftrightarrow\frac{6\left(x+33\right)}{30}=\frac{13x-14}{30}\)
\(\Leftrightarrow6x+198=13x-14\)
\(\Leftrightarrow6x-13x=-14-198\)
\(\Leftrightarrow-7x=-212\)
\(\Leftrightarrow x=\frac{212}{7}\)