Tìm x biết
A, x^2-4=8(x-2)
B,x^2-4x+4=9(x-2)
C,4x^2-12x+9=(5-x)^2
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NV
0
22 tháng 6 2018
\(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-8x+16=4\)
\(\Leftrightarrow\left(x-2\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)
Vậy...
22 tháng 6 2018
\(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow x^2-13x+22=0\)
\(\Leftrightarrow\left(x+\frac{13}{2}\right)^2=\frac{81}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{-21}{2}\end{cases}}\)
Vậy...
NV
3
BB
8 tháng 7 2016
x²-4=8(x-2)
=> x²-4=8x-16
=> x²-8x+16-4=0
=> (x-4)²-4=0
=>(x-4-2)(x-4+2)=0
=> (x-2)(x-6)=0
=> x-2=0 nên x=2
x-6 =0 nên x=6
8 tháng 7 2016
a) x^2 - 4 = 8(x - 2)
<=> (x - 2)(x + 2) - 8(x - 2) = 0
<=> (x - 2)(x+2-8)=0
<=>(x-2)(x-6)=0
<=>x-2=0 hoặc x-6=0
<=>x=2 hoặc x=6
Vậy S={2;6}
b)x^2-4x+4=9(x-2)
<=>(x-2)^2-9(x-2)=0
<=>(x-2)(x-2-9)=0
<=>(x-2)(x-11)=0
<=>x-2=0 hoặc x-11=0
<=>x=2 hoặc x=11
Vậy S={2;11}
c)4x^2-12x+9=(5-x)^2
<=>(2x)^2-2.2x.3+3^2=(5-x)^2
<=>(2x-3)^2-(5-x)^2=0
<=>(2x-3-5+x)(2x-3+5-x)=0
<=>(3x-8)(x+2)=0
<=>3x-8=0 hoặc x+2=0
<=>3x=8 hoặc x= - 2
<=>x=8:3(8 phần 3) hoặc x= -2
Vậy S={8:3 ; -2}
26 tháng 9 2021
e: 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 \(x=-\dfrac{7}{2}\)
f: Ta có: \(x^3-6x^2+12x-19=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)
\(\Leftrightarrow\left(x-2\right)^3=11\)
hay \(x=\sqrt[3]{11}+2\)
31 tháng 10 2021
\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
31 tháng 10 2021
a.
4x(x-5) - (x-1)(4x-3)-5=0
4x^2-20x-4x^2+3x+4x+3=0
(4x^2-4x^2)+(-20x+3x+4x)+3=0
13x+3 = 0
13x=-3
x=-3/13
b,
(3x-4)(x-2)-3x(x-9)+3=0
3x^2-6x-4x+8 - 3x^2+27x+3=0
(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0
17x+11=0
17x=-11
x=-11/17
c, 2(x+3)-x^2-3x=0
2(x+3) - x(x+3)=0
(x+3)(2-x)=0
TH1: x+3 = 0; x=-3
TH2: 2-x=0;x=2
HP
2 tháng 2 2021
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
HP
2 tháng 2 2021
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
24 tháng 8 2018
a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow\left(x^3-3x^2+3x-1\right)+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x-1+3x^2+2x+1=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x+3x^2+2x-x^3=1-1+8\)
\(\Leftrightarrow5x=8\)
\(\Leftrightarrow x=\dfrac{8}{5}\)
Vậy \(S=\left\{\dfrac{8}{5}\right\}\)
b) \(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)-8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-8\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow x-2=0\) hoặc \(x-6=0\)
:) \(x-2=0\Leftrightarrow x=2\)
:) \(x-6=0\Leftrightarrow x=6\)
Vậy \(S=\left\{2;6\right\}\)
c) \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)-9\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow x-2=0\) hoặc \(x-11=0\)
:) \(x-2=0\Leftrightarrow x=2\)
:) \(x-11=0\Leftrightarrow x=11\)
Vậy \(S=\left\{2;11\right\}\)
(d ko bít lèm)
#IDOL
4 tháng 8 2019
a,\(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow-3x^2+3x-1+3x^2+6x+3=8\)
\(\Leftrightarrow9x=6\)
\(\Leftrightarrow x=\frac{2}{3}\)
b,\(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-4=8x-16\)
\(\Leftrightarrow x^2+12x-8x=0\)
\(\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow x\left(x-2\right)-6\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
c,\(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow x^2-4x+4=9x-18\)
\(\Leftrightarrow x^2-4x+4-9x+18=0\)
\(\Leftrightarrow x^2-13x+22=0\)
\(\Leftrightarrow x^2-2x-11x+22=0\)
\(\Leftrightarrow x\left(x-2\right)-11\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
d,\(4x^2-12x+9=\left(5-x\right)^2\)
\(\Leftrightarrow4x^2-12x+9=25-10x+x^2\)
\(\Leftrightarrow4x^2-12x+9-25+10-x^2=0\)
\(\Leftrightarrow3x^2-2x-16=0\)
\(\Leftrightarrow3x^2+6x-8x-16=0\)
\(\Leftrightarrow3x\left(x+2\right)-8\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{8}{3}\end{matrix}\right.\)
5 tháng 8 2021
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
a, x^2-4=8(x-2)
=> x^2 - 4 = 8.x - 16
=> x^2 = (8.x - 16) - 4
=> x^2 = 8.x - (16+4)
=> x^2 = 8.x - 20
A, \(x^2-4=8\left(x-2\right)\)=> \(\left(x-2\right).\left(x+2\right)=8\left(x-2\right)=>\left(x-2\right).\left(x+2\right)-8\left(x-2\right)=0\)
=>\(\left(x-2\right).\left(x-6\right)=0\)
=> x = 2 hoặc x =6
B. \(x^2-4x+4=9\left(x-2\right)\)=> \(\left(x-2\right)^2=9\left(x-2\right)=>\left(x-2\right)^2-9\left(x-2\right)=0\)
=>\(\left(x-2\right).\left(x-11\right)=0\)=> x =2 hoặc x =11
C. \(4x^2-12x+9=\left(5-x\right)^2=>\left(2x-3\right)^2=\left(5-x\right)^2\)
=>\(\left(2x-3\right)^2-\left(5-x\right)^2=>\left(3x-8\right).\left(x+2\right)=0\)
=> x = 3/8 hoặc x = - 2