Bài 4: Phương trình tích

=>19x=7

=>x=7/19

\(19x-7=0\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\dfrac{7}{19}\)

\(\dfrac{x^2-x-6}{x-3}\\ =\dfrac{x^2-3x+2x-6}{x-3}\\ =\dfrac{x\left(x-3\right)+2\left(x-3\right)}{x-3}\\ =\dfrac{\left(x-3\right)\left(x+2\right)}{x-3}\\ =x+2\)

a,5x-20=0

<=>5x=20

<=>x=4

Vậy ....

b,<=>(x+5)(x-\(\dfrac{1}{2}\))=0

<=> x=-5 hoặc x=\(\dfrac{1}{2}\)

Vậy ....

d,<=>\(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}-10=0\)

<=>\(\dfrac{x-90-10}{10}+\dfrac{x-76-24}{12}+\dfrac{x-58-42}{14}+\dfrac{x-36-64}{16}=0\)

<=>(x-100)(\(\dfrac{1}{10}+\dfrac{1}{16}+\dfrac{1}{14}+\dfrac{1}{12}\))=0

<=>x=100

Vậy....

a: =>5x=20

=>x=4

b: =>(x-1/2)(x+5)=0

=>x=1/2 hoặc x=-5

c: \(\Leftrightarrow\dfrac{2\left(x+5\right)}{6\left(x-2\right)}-\dfrac{3\left(x-2\right)}{6\left(x-2\right)}=\dfrac{3\left(2x-3\right)}{6\left(x-2\right)}\)

=>6x-9=2x+10-3x+6=-x+16

=>7x=25

=>x=25/7

d: \(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)=0\)

=>x-100=0

=>x=100

a.

\(\left(x^2-x+1\right)\left(x^2-x+2\right)=12\)

Đặt \(x^2-x+1=y\) ta được:

\(y\left(y+1\right)=12\)

\(\Leftrightarrow y^2+y-12=0\)

\(\Leftrightarrow y^2+4y-3y-12=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=3\\x^2-x+1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-x+5=0\left(vn\right)\end{matrix}\right.\)

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

b.

\(3y^3-7y^2-7y+3=0\)

\(\Leftrightarrow3\left(y^3+1\right)-7y\left(y+1\right)=0\)

\(\Leftrightarrow3\left(y+1\right)\left(y^2-y+1\right)-7y\left(y+1\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y^2-3y+3-7y\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y^2-10y+3\right)=0\)

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

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

c.

\(x^2+2y^2=4x+4y-6=0\)

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

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

Do \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\) với mọi x;y

\(\Rightarrow\left(x-2\right)^2+2\left(y-1\right)^2\ge0\) ; \(\forall x;y\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x-2=0\\y-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

giúp mik bài hình vs e bài 1

e: =>(x+2)(3-4x)-(x+2)^2=0

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

=>(x+2)(-5x+1)=0

=>x=-2 hoặc x=1/5

Giải : 

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

\(2,2x^3+5x^2-3x=0\\ \Leftrightarrow2x^3+6x^2-x^2-3x=0\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(2x^2-x\right)\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\(3,6x^2-17x+12=0\\ \Leftrightarrow6x^2-8x-9x+12=0\\ \Leftrightarrow2x\left(3x-4\right)-3\left(3x-4\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(3x-4\right)=0\\\Leftrightarrow \left[{}\begin{matrix}2x-3=0\\3x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)

1)

`(x+1)^2 =4(x^2 -2x+1)`

<=> x^2 +2x+1=4x^2 -8x+4`

`<=> 4x^2 -x^2 -8x -2x +4 -1 =0`

`<=> 3x^2 -10x +3=0`

`<=> 3x^2 -9x-x+3=0`

`<=> 3x(x-3)-(x-3)=0`

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

\(< =>\left[{}\begin{matrix}x-3=0\\3x-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

2)

`2x^3 +5x^2 -3x=0`

`<=> x(2x^2 +5x-3)=0`

`<=> x(2x^2 +6x-x-3)=0`

`<=> x[2x(x+3)-(x+3)]=0`

`<=> x(x+3)(2x-1)=0`

\(< =>\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

3)

`6x^2 -17x+12=0`

`<=> 6x^2 -9x-8x+12=0`

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

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

\(< =>\left[{}\begin{matrix}2x-3=0\\3x-4=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)

\(1)\left(x+1\right)^2=4\left(x^2-2x+1\right)\)

\(\Leftrightarrow x^2+2x+1=4x^2-8x+4\)

\(\Leftrightarrow x^2+2x+1-4x^2+8x-4=0\)

\(\Leftrightarrow10x-3x^2-3=0\)

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

\(\Leftrightarrow-3x^2+9x+x-3=0\)

\(\Leftrightarrow\left(-3x^2+9x\right)+\left(x-3\right)=0\)

\(\Leftrightarrow3x\left(x-3\right)+\left(x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\3x=-1\end{matrix}\right.\)

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

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{3;\dfrac{-1}{3}\right\}\)

\(2)2x^3+5x^2-3x=0\)

\(\Leftrightarrow2x^3+6x^2-x^2-3x=0\)

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

\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{-3;0;\dfrac{1}{2}\right\}\)

\(3)6x^2-17x+12=0\)

\(\Leftrightarrow6x^2-8x-9x+12=0\)

\(\Leftrightarrow\left(6x^2-8x\right)-\left(9x-12\right)=0\)

\(\Leftrightarrow2x\left(3x-4\right)-3\left(3x-4\right)=0\)

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

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

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{4}{3};\dfrac{3}{2}\right\}\)

\(\left(x+1\right)^2=4\left(x^2-2x+1\right)\)

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

\(\Leftrightarrow x^2+2x+1-4x^2+8x-4=0\)

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

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

Vậy \(S=\left\{\dfrac{1}{3};3\right\}\)

 `(x+1)^2=4(x^2-2x+1)`

`<=>(x+1)^2-[2(x-1)]^2=0`

`<=>(x+1-2x+2)(x+1+2x-2)=0`

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

`<=>-x+3=0` hoặc `3x-1=0`

`<=>x=3` hoặc `x=1/3`

Vậy `S={1/3;3}`

Cần gấp ^^

\(\Leftrightarrow\left(\dfrac{x-11}{111}+1\right)+\left(\dfrac{x-12}{112}+1\right)=\left(\dfrac{x-23}{123}+1\right)+\left(\dfrac{x-24}{124}+1\right)\)

=>x+100=0

=>x=-100

a: =>\(\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+8}{92}+1\right)+\left(\dfrac{x+3}{97}+1\right)=0\)

=>x+100=0

=>x=-100

b: =>(x-11/111+1)+(x-12/112+1)=(x-23/123+1)+(x-24/124+1)

=>x+100=0

=>x=-100

=>x^2-4x+4+5=x^2+x-6

=>-4x+9=x-6

=>-5x=-15

=>x=3