Tìm x:
a)2.(5+3.x)+x=31
b)18-2.|x-6|=3^2+2020^0
c)3.x-1/11=x/4
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a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
\(a,\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\\ \Leftrightarrow2\left(x-3\right)=0\\ \Leftrightarrow x=3\)
\(b,4x^2-9=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(c,x^2+6x+9=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)
a. \(\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\)
\(\Leftrightarrow2\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\\ \Rightarrow x=-2\\ b,\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\\ \Rightarrow\left(2021x-1\right)\left(x-2020\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
a) \(\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\)
\(\Rightarrow2x=-4\Rightarrow x=-2\)
b) \(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)
\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
a: =>x*7/4+3/2=-4/5
=>x*7/4=-4/5-3/2=-8/10-15/10=-23/10
=>x=-23/10:7/4=-23/10*4/7=-92/70=-46/35
b: =>x*9/20=1/7+1/8=15/56
=>x=15/56:9/20=15/56*20/9=25/42
c: |x|=3,5
=>x=3,5 hoặc x=-3,5
d: |x|=-2,7
=>x thuộc rỗng
e: =>|x-1|=3-0,73=2,27
=>x-1=2,27 hoặc x-1=-2,27
=>x=-1,27 hoặc x=3,27
f: \(\Leftrightarrow7\cdot11x+11=0\)
=>77x=-11
=>x=-1/7
l: =>|x+3/4|=-2+5=3
=>x+3/4=3 hoặc x+3/4=-3
=>x=-15/4 hoặc x=9/4
a) \(x^2-x+x=4\)
\(x^2=4\)
\(x=\pm2\)
b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)
\(\left(x-5\right)\left(3x-2\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c) Ta có: \(a+b+c=5-3-2=0\)
\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)
d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :
\(t^2-11t+18=0\)
\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)
\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)
\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)
a) 2(5+3x)+x=31
=>10+6x+x=31
=>10+7x=31
=>7x=21
=>x=3
b)18-2.Ix-6I=3^2+2020^0
=>18-2.Ix+6I=9+1
=>2.Ix+6I=18-9+1
=> 2.Ix+6I=8
=> Ix+6I=4
=> x+6=4
x+6=-4
=> x= -2
x= -10
c) 3x-1/11=x/4
Quy đồng các số vs mẫu là 44
=> 12x/44-4/44=11x/44
Triệt tiêu mẫu
=> 12x-4=11x
=> 12x-11x=4
=. x=4
Thanks hưng nha