Tìm x:
a) (2x-1)2-(x+3)2=0
b) 2.(x+5)-x2-5x=0
K
Khách
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Những câu hỏi liên quan
7 tháng 9 2021
a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)
\(\Leftrightarrow x^3-x^3-1=x\)
hay x=-1
c: Ta có: \(56x^4+7x=0\)
\(\Leftrightarrow7x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d: Ta có: \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 12 2021
a.
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
b.
\(6x^2-7x+2=0\)
\(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow3x\left(2x-1\right)-2\left(2x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
C
1
31 tháng 10 2021
\(a,\Leftrightarrow\left(x-2\right)\left(5x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{5}\end{matrix}\right.\\ b,\Leftrightarrow2x^2+2x-x^2+4x-4-6=0\\ \Leftrightarrow x^2+6x-10=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{19}\\x=-3-\sqrt{19}\end{matrix}\right.\\ c,\Leftrightarrow2x^2-2x+9x-9=0\\ \Leftrightarrow\left(2x+9\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{2}\\x=1\end{matrix}\right.\)
C
1
17 tháng 8 2021
a: Ta có: \(40x^4+5x=0\)
\(\Leftrightarrow5x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(8x^2-2x-1=0\)
\(\Leftrightarrow8x^2-4x+2x-1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
31 tháng 7 2021
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
31 tháng 7 2021
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
4 tháng 9 2021
a: Ta có: \(2x^3-18x=0\)
\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)
\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)
\(\Leftrightarrow-13x=13\)
hay x=-1
c: Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8=3-3x^2\)
\(\Leftrightarrow3x=12\)
hay x=4
4 tháng 9 2021
a) 2x3-18x=0
⇔ 2x(x2-9)=0
⇔ 2x(x-3)(x+3)=0
⇔ \(\left\{{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b)(3x-1)(2x+1)-6x(x+2)=11
⇔ 6x2+x-1-6x2-12x=11
⇔ -11x=12
\(\Leftrightarrow x=-\dfrac{12}{11}\)
c) (x-1)3-(x+2).(x2-2x+4)=3.(1-x2)
⇔ x3-3x2+3x-1-x3-8-3+3x2=0
⇔ 3x=12
⇔ x=4
25 tháng 12 2021
a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
TP
5
4 tháng 9 2021
a: \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow x^2+7x-x^2-x+6=0\)
hay x=-1
b: Ta có: \(\left(x+2\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
KK
4 tháng 9 2021
b. (x + 2)2 - x2 + 4 = 0
<=> (x + 2 - x)(x + 2 + x) + 4 = 0
<=> 2(2 + 2x) + 4 = 0
<=> 4(1 + x) + 4 = 0
<=> 4(1 + x) = -4
<=> 1 + x = -1
<=> x = -1 - 1
<=> x = -2
Câu a) (2X-1)2-(X+3)2=0
(2X-1-X-3)(2X-1+X+3)=0
(X-4)(3X+2)=0
-> X-4=0 HOẶC 3X+2=0
xét:X-4=0
X=0+4=4
Xét:3X+2=0
3X=0-2=-2
X= -2:3= -2/3
=>X=4 HOẶC -2/3
CÂU b) 2(X+5)-X2-5X=0
2(X+5)-(x2+5X)=0
2(X+5)-X(x+5)=0
(X+5)(2-x)=0
-> X+5=0 HOẶC 2-X=0
XÉT: X+5=0 XÉT: 2-X=0
X=0-5= -5 X=2-0=2
=>X= -5 HOẶC 2