Tìm x:
-5x^2+16x-3=0
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14 tháng 7 2023
a: =>2^4x<2^28
=>4x<28
=>x<7
b: =>5^3x+3<5
=>3x+3<1
=>3x<-2
=>x<-2/3
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JH
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KH
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18 tháng 8 2021
a, \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left(3x+3\right)^2=0\Leftrightarrow\left(4x-3x-3\right)\left(4x+2x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)
b, \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow x=-2;x=\frac{1}{3}\)
c, \(5x^3-20x=0\Leftrightarrow5x\left(x^2-4\right)=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow x=0;x=\pm2\)
JH
1
18 tháng 8 2021
1: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
2: Ta có: \(\left(5x-4\right)^2-49x^2=0\)
\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)
\(\Leftrightarrow\left(2x+4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3: Ta có: \(5x^3-20x=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
11 tháng 8 2019
a) 3x4 - 13x3 + 16x2 - 13x + 3 = 0
(x - 3)(3x - 1)(x2 - x + 1) = 0
nhưng vì x2 - x + 1 # 0 nên:
x - 3 = 0 hoặc 3x - 1 = 0
x = 0 + 3 3x = 0 + 1
x = 3 3x = 1
x = 1/3
b) 6x4 + 5x3 - 38x2 + 5x + 6 = 0
(x - 2)(x + 3)(3x + 1)(2x - 1) = 0
x - 2 = 0 hoặc x + 3 = 0 hoặc 3x + 1 = 0 hoặc 2x - 1 = 0
x = 0 + 2 x = 0 - 3 3x = 0 - 1 2x = 0 + 1
x = 2 x = -3 3x = -1 2x = 1
x = -1/3 x = 1/2
19 tháng 6 2021
a, 4x2 - 49 = 0
⇔⇔ (2x)2 - 72 = 0
⇔⇔ (2x - 7)(2x + 7) = 0
⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72
b, x2 + 36 = 12x
⇔⇔ x2 + 36 - 12x = 0
⇔⇔ x2 - 2.x.6 + 62 = 0
⇔⇔ (x - 6)2 = 0
⇔⇔ x = 6
19 tháng 6 2021
e, (x - 2)2 - 16 = 0
⇔⇔ (x - 2)2 - 42 = 0
⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0
⇔⇔ (x - 6)(x + 2) = 0
⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2
f, x2 - 5x -14 = 0
⇔⇔ x2 + 2x - 7x -14 = 0
⇔⇔ x(x + 2) - 7(x + 2) = 0
⇔⇔ (x + 2)(x - 7) = 0
⇔{x+2=0x−7=0⇔{x=−2x=7
25 tháng 9 2021
\(\left(5x-4\right)^2-16x^2=0\)
\(\Leftrightarrow\left(5x-4\right)^2-\left(4x\right)^2=0\)
\(\Leftrightarrow\left(5x-4-4x\right)\left(5x-4+4x\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(9x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{9}\end{matrix}\right.\)
25 tháng 9 2021
\(\left(5x-4\right)^2-16x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(4x\right)^2=0\\ \Leftrightarrow\left(5x-4-4x\right).\left(5x-4+4x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x-4-4x=0\\5x-4+4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{9}\end{matrix}\right.\\ \Rightarrow S=\left\{\dfrac{4}{9};4\right\}\)
21 tháng 10 2018
\(5x^2+2y^2-6xy+16x-8y+16=0\)
\(\Rightarrow10x^2+4y^2-12xy+32x-16y+32=0\)
\(\Rightarrow\left(9x^2-12xy+4y^2\right)+\left(24x-16y\right)+16+\left(x^2+8x+16\right)=0\)
\(\Rightarrow\left(3x-2y\right)^2+2.\left(3x-2y\right).4+4^2+\left(x+4\right)^2=0\)
\(\Rightarrow\left(3x-2y+4\right)^2+\left(x+4\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}3x-2y+4=0\\x+4=0\end{cases}\Rightarrow}\hept{\begin{cases}-12-2y+4=0\\x=-4\end{cases}\Rightarrow\hept{\begin{cases}y=-4\\x=-4\end{cases}}}\)
Vậy \(x=y=-4\)
\(-5x^2+16x-3=0\)
\(-5x^2+x+15x-3=0\)
\(x\cdot\left(-5x+1\right)-3\cdot\left(-5x-1\right)=0\)
\(\left(-5x-1\right)\cdot\left(x-3\right)=0\)
\(\hept{\begin{cases}-5x-1=0\\x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{1}{5}\\x=3\end{cases}}}\)
Vậy.......
-5x^2 +15x +x-3=0
5x(-x +3)-(-x+3)=
(5x-1)(-x+3)=0
5x-1 =0 hoặc -x+3=0
x=1/5 hoặc x=3