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21 tháng 11 2018

Ta có: 5( x + 2 ) - x2- 2x = 0

=> 5x + 10 - x2 - 2x = 0

=> x2 - 3x + 10 = 0

=> (x2 + 2x) - (5x - 10 ) = 0

=> (x-5)(x-2)=0

=> x = 5 hoặc x = 2.

21 tháng 11 2018

\(5\left(x+2\right)-x^2-2x=0\)

\(\Leftrightarrow5x+10-x^2-2x=0\)

\(\Leftrightarrow-x^2+3x+10=0\)

\(\Leftrightarrow-x^2-2x+5x+10=0\)

\(\Leftrightarrow-x\left(x+2\right)+5\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\5-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=5\end{cases}}}\)

a: Ta có: \(2x\left(x-3\right)+x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)

Câu 1: 

Ta có: \(x^2-5x+4=0\)

\(\Leftrightarrow x^2-x-4x+4=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

Vậy: S={1;4}

Câu 2: 

Ta có: \(3x^2-7x+3=0\)

\(\Delta=\left(-7\right)^2-4\cdot3\cdot3=49-36=13\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{13}}{6}\\x_2=\dfrac{7+\sqrt{13}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7-\sqrt{13}}{6};\dfrac{7+\sqrt{13}}{6}\right\}\)

Câu 3: 

Ta có: \(5x^2-x-4=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{4}{5}\right\}\)

Câu 4: 

Ta có: \(7x^2+x-8=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{8}{7}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{8}{7}\right\}\)

Câu 1x^2-5x+4=0

<=>(x-1)(x-4)=0

<=>[x=1;x=4

Câu 2 3x^2-7x+3=0

x=7/6-căn bậc hai(13)/6, x=căn bậc hai(13)/6+7/6

x=7/6-căn bậc hai(13)/6, x=căn bậc hai(13)/6+7/6

Câu 3 5*x^2 -x-4 = 0

x=-4/5, x=1

Câu 4 7*x^2 +x-8 = 0

x=-8/7, x=1

bn ơi mk giải thế có chỗ nào ko hiểu bn có thể hỏi mk nhé

 

 
10 tháng 10 2021

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

10 tháng 10 2021

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

9 tháng 10 2021

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

9 tháng 10 2021

\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\)

\(\Rightarrow x^3-25x-x^3-8=17\)

\(\Rightarrow25x=-25\Rightarrow x=-1\)

27 tháng 8 2021

`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`

27 tháng 8 2021


`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`