tím x biết:
(2x+1)^2-4*(x+2)=9
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\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
\(=>4.\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)
\(=>4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(=>4x-13=11\)\(=>4x=11+13=24\)
\(=>x=24:4=6\)
CHúc bạn Hk tốt!!!
1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
\(\left(2x+1\right)^2-4.\left(x+2\right)=9\)
\(\Leftrightarrow4x^2+4x+1-4x-8=9\)
\(\Leftrightarrow4x^2=16\)
\(\Leftrightarrow x^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
vậy...............
\(\left(2x+1\right)^2-4\left(x+2\right)=9\)
\(\Leftrightarrow4x^2+4x+1-4x-8=9\)
\(\Leftrightarrow4x^2-7-9=0\)
\(\Leftrightarrow4x^2-16=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)