Tìm x:
a)2x.(x+3)+2x.(6-x)=7
b)4.(x+1)-2.(x-3)-3.(x-4)=0
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a: Ta có: \(x\left(2-x\right)+x^2+x=7\)
\(\Leftrightarrow2x-x^2+x^2+x=7\)
\(\Leftrightarrow3x=7\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)
\(\Leftrightarrow2x-x^2+x^2+x=7\)
\(\Leftrightarrow3x=7\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)
\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)
\(\Leftrightarrow9x^2=16\)
\(\Leftrightarrow x^2=\dfrac{16}{9}\)
hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)
a) \(=2x^2-7x-15-2x^2+6x+x+7=-8\)
b) \(=2x^2+x-x^3-2x^2+x^3-x+3=3\)
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: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
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: =>x+5>0
hay x>-5
b: =>2x+1<0
hay x<-1/2
c: =>(x-1)(x-4)>0
=>x>4 hoặc x<1