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8 tháng 8 2018
a) \(x+x^2=0\Leftrightarrow x\left(1+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
b) \(x+1-\left(x+1\right)^2=0\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow-x\left(x+1\right)\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
c) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\Leftrightarrow\left(15y-3\right)\left(4y-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{15}=\dfrac{1}{5}\\x=\dfrac{9}{4}\end{matrix}\right.\)
d) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\Leftrightarrow\left(8-27z\right)\left(25z+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}z=\dfrac{8}{27}\\z=\dfrac{-7}{25}\end{matrix}\right.\)
22 tháng 10 2020
a) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\)
\(\Leftrightarrow\left(15y-3\right)\left(4y-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}15y-3=0\\4y-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{5}\\y=\frac{9}{4}\end{matrix}\right.\)
Vây \(y\in\left\{\frac{1}{5};\frac{9}{4}\right\}\)
b) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\)
\(\Leftrightarrow\left(8-27z\right)\left(25z+7\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}z=\frac{8}{27}\\z=-\frac{7}{25}\end{matrix}\right.\)
Vậy \(z\in\left\{\frac{8}{27};-\frac{7}{25}\right\}\)
c) \(13y\left(y-8\right)-2y+16=0\)
\(\Leftrightarrow13y\left(y-8\right)-2\left(y-8\right)=0\)
\(\Leftrightarrow\left(13y-2\right)\left(y-8\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{2}{13}\\y=8\end{matrix}\right.\)
Vậy \(y\in\left\{\frac{2}{13};8\right\}\)
d) \(-10y\left(y+2\right)-y-2=0\)
\(\Leftrightarrow\left(-10y-1\right)\left(y+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=-\frac{1}{10}\end{matrix}\right.\)
Vậy \(y\in\left\{-2;-\frac{1}{10}\right\}\)
e) \(x\left(x+19\right)^2-\left(x+19\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+19\right)^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-19\end{matrix}\right.\)
Vậy \(x\in\left\{1;-19\right\}\)
TG
14 tháng 1 2021
a) (x - 7)(2x + 8) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\2x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy: S = {7; -4}
b) Tương tự câu a
c) (x - 1)(2x + 7)(x2 + 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\\x^2+2=0\end{matrix}\right.\)
Mà: x2 + 2 > 0 với mọi x
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{2}\right\}\)
d) (2x - 1)(x + 8)(x - 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-8\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};-8;5\right\}\)
TT
14 tháng 1 2021
a/ Pt \(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy \(S=\left\{7;-4\right\}\)
b/ pt \(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\5x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)
c/ pt \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\) (\(x^2+2>0\forall x\))\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
d/ pt \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)
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 \(S=\left\{2;2\right\}\)
D
1
25 tháng 10 2021
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(x^2-1=0\Rightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
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: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
x + x2 = 0
=> x(1 + x) = 0
=> x = 0 hoặc x + 1 = 0
=> x = 0 hoặc x = -1
vậy_
mk biến đổi về pt tích, sau đó bạn tính nốt nhé:
b) \(x+1-\left(x+1\right)^2=0\)
<=> \(\left(x+1\right)\left(1-x-1\right)=0\)
<=> \(-x\left(x+1\right)=0\)
c) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\)
<=> \(3\left(4y-9\right)\left(5y-1\right)=0\)
d) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\)
<=> \(\left(25z+7\right)\left(8-27z\right)=0\)