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T
7 tháng 1 2019
a)\(4x^3-9x=0\Leftrightarrow x\left(4x^2-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=\frac{9}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)
Vậy x = 0 hoặc \(x=\frac{3}{2}\)
b) \(x^3+8x=0\Leftrightarrow x\left(x^2+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=-8\left(L\right)\end{cases}}\)
Vậy x = 0
c) \(-x^3+9x=0\Leftrightarrow x\left(-x^2+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x^2+9=0\\x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=9\\x=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=0\end{cases}}\)
Vậy ...
19 tháng 8 2020
a,Cách 1 : \(x^2-10x+9=0\Leftrightarrow\left(x-1\right)\left(x-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=9\end{cases}}\)
Cách 2 : Dung p^2 nhẩm nghiệm p^2 bậc 2 vì : 1 - 10 + 9 = 0
\(\Leftrightarrow\orbr{\begin{cases}x_1=1\\x_2=\frac{c}{a}=9\end{cases}}\)
b, Cách 1 : \(8x^2-2x-15=0\Leftrightarrow\left(4x+5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=\frac{3}{2}\end{cases}}\)
Cách 2 : \(\Delta=\left(-2\right)^2-4.8.\left(-15\right)=484>0\)
Pp có 2 nghiệm phân biệt : \(x_1=\frac{-2-\sqrt{484}}{16};x_2=\frac{-2+\sqrt{484}}{16}\)
PN
20 tháng 8 2020
toán 9 à bạn ?
c,\(2x^2+8x-7=0\)
Ta có : \(\Delta=8^2-4.\left(-7\right).2=64+56=120\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-8+\sqrt{120}}{4}=-2+\frac{\sqrt{120}}{4}\\x=\frac{-8-\sqrt{120}}{4}=-2-\frac{\sqrt{120}}{4}\end{cases}}\)
d,\(3x^2-15x+3=0\)
Ta có : \(\Delta=\left(-15\right)^2-4.3.3=225-36=189\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{15+\sqrt{189}}{6}\\x=\frac{15-\sqrt{189}}{6}\end{cases}}\)
e,\(16x^2-24x-4=0\Leftrightarrow4x^2-6x-1=0\)
Ta có : \(\Delta=\left(-6\right)^2-4.4.\left(-1\right)=36+16=52\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{6+\sqrt{52}}{8}\\x=\frac{6-\sqrt{52}}{8}\end{cases}}\)
f, \(-5x^2+6x+3=0\)
Ta có : \(\Delta=6^2-4.3.\left(-5\right)=36+60=96\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-6+\sqrt{96}}{-10}\\x=\frac{-6-\sqrt{96}}{-10}\end{cases}}\)
i, \(6x^2-9x+40=0\)
Ta có : \(\Delta=\left(-9\right)^2-4.6.40=81-960=-879\)
do đen ta < 0 => vô nghiệm
18 tháng 8 2020
a)
Cách 1:
Ta có: \(x^2-10x+9=0\)
\(\Leftrightarrow x^2-x-9x+9=0\)
\(\Leftrightarrow x\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
Vậy: S={1;9}
Cách 2:
Ta có: \(x^2-10x+9=0\)
\(\Leftrightarrow x^2-10x+25-16=0\)
\(\Leftrightarrow\left(x-5\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy: S={9;1}
b)
Cách 1:
Ta có: \(8x^2-2x-15=0\)
\(\Leftrightarrow8x^2-12x+10x-15=0\)
\(\Leftrightarrow4x\left(2x-3\right)+5\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{-5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{3}{2};\frac{-5}{4}\right\}\)
Cách 2:
Ta có: \(8x^2-2x-15=0\)
\(\Leftrightarrow8\left(x^2-\frac{1}{4}x-\frac{15}{8}\right)=0\)
\(\Leftrightarrow x^2-\frac{1}{4}x-\frac{15}{8}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{1}{8}+\frac{1}{64}-\frac{121}{64}=0\)
\(\Leftrightarrow\left(x-\frac{1}{8}\right)^2=\frac{121}{64}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{8}=\frac{11}{8}\\x-\frac{1}{8}=-\frac{11}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{12}{8}=\frac{3}{2}\\x=\frac{-11+1}{8}=\frac{-10}{8}=\frac{-5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{3}{2};\frac{-5}{4}\right\}\)
c) Ta có: \(2x^2+8x-7=0\)
\(\Leftrightarrow2\left(x^2+4x-\frac{7}{2}\right)=0\)
\(\Leftrightarrow x^2+4x+4-\frac{15}{2}=0\)
\(\Leftrightarrow\left(x+2\right)^2=\frac{15}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{\frac{15}{2}}\\x+2=-\sqrt{\frac{15}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\frac{15}{2}}-2\\x=-\sqrt{\frac{15}{2}}-2\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{\frac{15}{2}}-2;-\sqrt{\frac{15}{2}}-2\right\}\)
d) Ta có: \(3x^2-15x+3=0\)
\(\Leftrightarrow3\left(x^2-5x+1\right)=0\)
\(\Leftrightarrow x^2-5x+1=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{21}{4}=0\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\frac{21}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{5}{2}=\frac{\sqrt{21}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{21}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{21}+5}{2}\\x=\frac{-\sqrt{21}+5}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{\sqrt{21}+5}{2};\frac{-\sqrt{21}+5}{2}\right\}\)
e) Ta có: \(16x^2-24x-4=0\)
\(\Leftrightarrow4\left(4x^2-6x-1\right)=0\)
\(\Leftrightarrow4x^2-6x-1=0\)
\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot\frac{3}{2}+\frac{9}{4}-\frac{13}{4}=0\)
\(\Leftrightarrow\left(2x-\frac{3}{2}\right)^2=\frac{13}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{3}{2}=\frac{\sqrt{13}}{2}\\2x-\frac{3}{2}=-\frac{\sqrt{13}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{3+\sqrt{13}}{2}\\2x=\frac{3-\sqrt{13}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3+\sqrt{13}}{2}:2=\frac{3+\sqrt{13}}{4}\\x=\frac{3-\sqrt{13}}{2}:2=\frac{3-\sqrt{13}}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{3+\sqrt{13}}{4};\frac{3-\sqrt{13}}{4}\right\}\)
f) Ta có: \(-5x^2+6x+3=0\)
\(\Leftrightarrow-5\left(x^2-\frac{6}{5}x-\frac{3}{5}\right)=0\)
\(\Leftrightarrow x^2-\frac{6}{5}x-\frac{3}{5}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{5}+\frac{9}{25}-\frac{24}{25}=0\)
\(\Leftrightarrow\left(x-\frac{3}{5}\right)^2=\frac{24}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{5}=\frac{2\sqrt{6}}{5}\\x-\frac{3}{5}=\frac{-2\sqrt{6}}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3+2\sqrt{6}}{5}\\x=\frac{3-2\sqrt{6}}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{3+2\sqrt{6}}{5};\frac{3-2\sqrt{6}}{5}\right\}\)
i) Ta có: \(6x^2-9x+40=0\)
\(\Leftrightarrow6\left(x^2-\frac{3}{2}x+\frac{20}{3}\right)=0\)
\(\Leftrightarrow x^2-\frac{3}{2}x+\frac{20}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}+\frac{293}{48}=0\)
\(\Leftrightarrow\left(x-\frac{3}{4}\right)^2+\frac{293}{48}=0\)(vô lý)
Vậy: \(S=\varnothing\)
16 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
NH
5 tháng 3 2023
1)
`7x^2 -49x=0`
`<=>x(7x-49)=0`
\(< =>\left[{}\begin{matrix}x=0\\7x-49=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
2)
`8x^2 -16x=0`
`<=>x(8x-16)=0`
\(< =>\left[{}\begin{matrix}x=0\\8x-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
3)
`2x^3 +40x=0`
`<=>x(2x^2 +40)=0`
`<=>x=0` hoặc`2x^2 +40=0`
`<=>x=0` hoặc `2x^2 =-40` (vô lí vì `2x^2` luôn lớn hơn hoặc bằng 0)
`<=>x=0`
4)
`-x^3 +16x=0`
`<=>x^3 -16x=0`
`<=>x(x^2 -16)=0`
\(< =>\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
mấy cái kia cũng làm giống vậy
1)\(x^2-x=x\left(x-1\right)=0\)
\(\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)