a/ 2.(1/2x -1/3)-3/2=1/4
b/ (1/2+2x)(2x-3)=0
c/ (4x-5)(5/4x-2)=0
d/ 1/4-(2x+1/2)mũ hai=0
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9 tháng 9 2021
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
KK
17 tháng 2 2021
sửa lại chút: a) (2x+1)^2-2x-1=2 b) (x^2-3x)^2+5(x^2-3x)+6=0 c) (x^2-x-1)(x^2-x)-2=0 d) (5-2x)^2+4x-10=8 e) (x^2+2x+3)(x^2+2x+1)=3 f) x(x-1)(x^2-x+1)-6=0
17 tháng 2 2021
a) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1}{2};-1\right\}\)
25 tháng 2 2022
1: \(\Leftrightarrow\left(x-4\right)^2+14=-9\left(x-4\right)\)
\(\Leftrightarrow x^2-8x+16+14+9x-36=0\)
\(\Leftrightarrow x^2+x-6=0\)
=>(x+3)(x-2)=0
=>x=-3(nhận) hoặc x=2(nhận)
2: \(\Leftrightarrow\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9=0\)
\(\Leftrightarrow16x^2-8x+2x-1-4x^2-2x-12x^2+9=0\)
=>-8x+8=0
hay x=1(nhận)
c: \(\dfrac{1}{2\left(x-3\right)}-\dfrac{3x-5}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{2}\)
\(\Leftrightarrow x-1-2\left(3x-5\right)=\left(x-3\right)\left(x-1\right)\)
\(\Leftrightarrow x^2-4x+3=x-1-6x+10=-5x+9\)
\(\Leftrightarrow x^2+x-6=0\)
=>(x+3)(x-2)=0
=>x=-3(nhận) hoặc x=2(nhận)
19 tháng 8 2021
a: Ta có: \(\left(3x+5\right)^2-4x^2=0\)
\(\Leftrightarrow\left(3x+5+2x\right)\left(3x+5-2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
23 tháng 10 2021
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
1 tháng 6 2018
Tìm x:
1. \(25x^2-20x+4=0\)
⇔ \(\left(5x-2\right)^2=0\)
⇔ \(5x-2=0\)
⇔ \(5x=2\)
⇔ \(x=\dfrac{2}{5}\)
⇒ S = \(\left\{\dfrac{2}{5}\right\}\)
2. \(\left(2x-3\right)^2-\left(2x+1\right).\left(2x-1\right)=0\)
⇔ \(4x^2-12x+9-\left(4x^2-1\right)=0\)
⇔ \(4x^2-12x+9-4x^2+1=0\)
⇔ \(-12x+10=0\)
⇔ \(-12x=-10\)
⇔ \(x=\dfrac{5}{6}\)
⇒ S \(=\left\{\dfrac{5}{6}\right\}\)
3. \(\left(\dfrac{1}{2}x-1\right)\left(\dfrac{1}{2}x+1\right)-\left(\dfrac{1}{2}x-1\right)^2=0\)
⇔ \(\dfrac{1}{4}x^2-1-\left(\dfrac{1}{4}x^2-x+1\right)=0\)
⇔ \(\dfrac{1}{4}x^2-1-\dfrac{1}{4}x^2+x-1=0\)
⇔ \(-2+x=0\)
⇔ \(x=2\)
⇒ S \(=\left\{2\right\}\)
4. \(\left(2x-3\right)^2+\left(2x+5\right)^2=8\left(x+1\right)^2\)
⇔ \(4x^2-12x+9+4x^2+20x+25=8\left(x^2+2x+1\right)\)
⇔ \(8x^2+8x+34=8x^2+16x+8\)
⇔ \(8x+34=16x+8\)
⇔ \(8x-16x=8-34\)
⇔ \(-8x=-26\)
⇔ \(x=\dfrac{13}{4}\)
⇒ S \(=\left\{\dfrac{13}{4}\right\}\)
5.\(4x^2+12x-7=0\)
⇔ \(4x^2+14x-2x-7=0\)
⇔ \(2x\left(2x+7\right)-\left(2x+7\right)=0\)
⇔ \(\left(2x+7\right)\left(2x-1\right)=0\)
⇔ \(\left[{}\begin{matrix}2x+7=0\\2x-1=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-7}{2};\dfrac{1}{2}\right\}\)
6. \(\dfrac{1}{4}x^2+\dfrac{2}{3}x-\dfrac{5}{9}=0\)
⇔ \(9x^2+24x-20=0\)
⇔ \(9x^2+30x-6x-20=0\)
⇔ \(3x\left(3x+10\right)-2\left(3x+10\right)=0\)
⇔ \(\left(3x+10\right)\left(3x-2\right)=0\)
⇔ \(\left[{}\begin{matrix}3x+10=0\\3x-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-10}{3};\dfrac{2}{3}\right\}\)
1 tháng 6 2018
7. \(24\dfrac{8}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)
⇔ \(\dfrac{224}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)
⇔ \(896-9x^2-12x=0\)
⇔ \(-896+9x^2+12x=0\)
⇔ \(9x^2+12x-896=0\)
⇔ \(9x^2-84x+96x-896=0\)
⇔ \(3x\left(3x-28\right)+32\left(3x-28\right)=0\)
⇔ \(\left(3x-28\right)\left(3x+32\right)=0\)
⇔ \(\left[{}\begin{matrix}3x-28=0\\3x+32=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=\dfrac{-32}{3}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-32}{3};\dfrac{28}{3}\right\}\)
N
10 tháng 10 2021
\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)
\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)
\(-18x+13=0\)
\(x=\dfrac{13}{18}\)
Vậy \(S=\left\{\dfrac{13}{18}\right\}\)
\(b.\left(x-1\right)^3-125=0\)
\(\left(x-1\right)^3=125\)
\(x-1=5\)
\(x=6\)
Vậy \(S=\left\{6\right\}\)
\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)
\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(S=\left\{1;-2\right\}\)
\(d.x^2-4x+4+x^2-2xy+y^2=0\)
\(\left(x-2\right)^2+\left(x-y\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy \(S=\left\{2;2\right\}\)
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