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\(x^2-9x+8=0\)

=>\(x^2-x-8x+8=0\)

=>x(x-1)-8(x-1)=0

=>(x-1)(x-8)=0

=>\(\left[{}\begin{matrix}x-1=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

13 tháng 12

x^2-9x+8=0

(x-8)(x-1)=0

x=8 hoặc x=1.

21 tháng 8 2021

undefinedMình trình bày trong hình ^^ Bn tham khảo nhé

d: Ta có: \(9x^2+6x-8=0\)

\(\Leftrightarrow9x^2+12x-6x-8=0\)

\(\Leftrightarrow\left(3x+4\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

e: Ta có: \(x\left(x-2\right)+x-2=0\)

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

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

f: Ta có: \(5x\left(x-3\right)-x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

20 tháng 5 2022

|9x−8|+|7x−6|+|5x−4|+|3x−2|+x=0(1)|9x−8|+|7x−6|+|5x−4|+|3x−2|+x=0(1).

Vì |9x−8|+|7x−6|+|5x−4|+|3x−2|>0∀x|9x−8|+|7x−6|+|5x−4|+|3x−2|>0∀x

Nên từ (1) ⇒x<0⇒9x−8;7x−6;5x−4;3x−2<0⇒x<0⇒9x−8;7x−6;5x−4;3x−2<0.

Phương trình (1) trở thành:

8−9x+6−7x+4−5x+2−3x+x=0⇔20−23x=0⇔x=20/23>0(ktm)

20 tháng 5 2022

Tham khảo vào đi .-.

8 tháng 8 2016

d) <=>x2-5x-x+5=0

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

<=>(x-5)(x-1)=0

<=>x=5 hoặc x=1

9 tháng 8 2016

thank nha

16 tháng 9 2017

9x2 +6x-8=0

<=> 9x2 +6x+1-9=0

<=> (3x+1)^2 - 3^2 = 0

<=> (3x+1-3)(3x+1+3)=0

<=> (3x-2)(3x+4)=0

=> TH1: 3x-2=0 <=> x=2/3

TH2: 3x+4=0 <=> x= -4/3

vậy ....................

16 tháng 9 2017

\(9x^2+6x-8=0\)

\(9x^2+6x+1-9=0\)

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

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

\(\left(3x-2\right)\left(3x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\3x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=2\\3x=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-4}{3}\end{cases}}\)

vay \(\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-4}{3}\end{cases}}\)

19 tháng 9 2015

Ta có :

\(9x^2-6x-8=0\)

\(x.\left(9x-6\right)=8\)

Lập bảng xét là xong ok 

17 tháng 11 2021

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

17 tháng 11 2021

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

NV
20 tháng 7 2021

a.

\(\Leftrightarrow\left(3x-1\right)^3=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow3x-1=-\dfrac{1}{2}\)

\(\Leftrightarrow3x=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{6}\)

b.

\(\Leftrightarrow\left(2x-1\right)\left(x-4\right)-x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x-1-x\right)=0\)

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

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

c.

\(\Leftrightarrow3x\left(5x-2\right)-2\left(5x-2\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(5x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)

31 tháng 10 2019

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

\(\Leftrightarrow3x\left(x-2\right)-\left(x-2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)

31 tháng 10 2019

\(B1:\)

\(3x\left(x-2\right)-\left(x-2\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)

19 tháng 10 2021

Th1: 9x+3=0

9x=-3

x= -1/3

Th2 : 2x-8=0

2x=8

x=4

19 tháng 10 2021

\(\Rightarrow3\left(3x+1\right).2\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=4\end{matrix}\right.\)

13 tháng 7 2019

\(x^3-9x-8=0\)

\(\Rightarrow xxx-9x-8=0\)

\(\Rightarrow\left(xx-9\right)x-8=0\)

\(\Rightarrow\left(x^2-9\right)x=0+8=8\)

\(\Rightarrow x=-1\)

13 tháng 7 2019

x3 - 9x - 8 = 0

=> x3 - x - 8x - 8 = 0

=> x(x2 - 1) - 8(x + 1) = 0

=> x(x - 1)(x + 1) - 8(x + 1) = 0

=> (x + 1)[x(x - 1) - 8) = 0 

=> x + 1 = 0 hoặc x(x - 1) - 8 = 0

=> x = -1 hoặc x(x - 1) = 8