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bài 1 câu a bỏ x= nhé !

b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)

\(\Leftrightarrow x^2+7x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

a: =>|x-7|=3-2x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)

b: =>|2x-3|=4x+9

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)

c: =>3x+5=2-5x hoặc 3x+5=5x-2

=>8x=-3 hoặc -2x=-7

=>x=-3/8 hoặc x=7/2

a) (x-1)(5x+3)=(3x-8)(x-1)

= (x-1)(5x+3)-(3x-8)(x-1)=0

=(x-1)[(5x+3)-(3x-8)]=0

=(x-1)(5x+3-3x+8)=0

=(x-1)(2x+11)=0

\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0

\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)

Vậy S={1;\(\dfrac{-11}{2}\)}

b) 3x(25x+15)-35(5x+3)=0

=3x.5(5x+3)-35(5x+3)=0

=15x(5x+3)-35(5x+3)=0

=(5x+3)(15x-35)=0

\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0

\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)

Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}

c) (2-3x)(x+11)=(3x-2)(2-5x)

=(2-3x)(x+11)-(3x-2)(2-5x)=0

=(3x-2)[(x+11)-(2-5x)]=0

=(3x-2)(x+11-2+5x)=0

=(3x-2)(6x+9)=0

\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0

\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)

Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}

d) (2x²+1)(4x-3)=(2x²+1)(x-12)

=(2x²+1)(4x-3)-(2x²+1)(x-12)=0

=(2x²+1)[(4x-3)-(x-12)=0

=(2x²+1)(4x-3-x+12)=0

=(2x²+1)(3x+9)=0

\(\Leftrightarrow\)2x²+1=0 hoặc 3x+9=0

\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3

Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}

e) (2x-1)²+(2-x)(2x-1)=0

=(2x-1)[(2x-1)+(2-x)=0

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

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

\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0

\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1

Vậy S={\(\dfrac{-1}{2}\);-1}

f)(x+2)(3-4x)=x²+4x+4

=(x+2)(3-4x)=(x+2)²

=(x+2)(3-4x)-(x+2)²=0

=(x+2)[(3-4x)-(x+2)]=0

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

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

\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0

\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)

Vậy S={-2;\(\dfrac{1}{5}\)}

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

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

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

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

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

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

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

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

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

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)

giải đc sao pn dễ mk

chẳng ai giải, thôi mình giải vậy!

a) Đặt \(y=x^2+4x+8\),phương trình có dạng:

\(t^2+3x\cdot t+2x^2=0\)

\(\Leftrightarrow t^2+xt+2xt+2x^2=0\)

\(\Leftrightarrow t\left(t+x\right)+2x\left(t+x\right)=0\)

\(\Leftrightarrow\left(2x+t\right)\left(t+x\right)=0\)

\(\Leftrightarrow\left(2x+x^2+4x+8\right)\left(x^2+4x+8+x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)vậy tập nghiệm của phương trình là:S={-2;-4}

b) nhân 2 vế của phương trình với 12 ta được:

\(\left(6x+7\right)^2\left(6x+8\right)\left(6x+6\right)=72\)

Đặt y=6x+7, ta được:\(y^2\left(y+1\right)\left(y-1\right)=72\)

giải tiếp ra ta sẽ được S={-2/3;-5/3}

c) \(\left(x-2\right)^4+\left(x-6\right)^4=82\)

S={3;5}

d)s={1}

e) S={1;-2;-1/2}

f) phương trình vô nghiệm

`a,` \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

`<=> (5(5x+2))/30 - (10(8x-1))/30 = (6(4x+2))/30 - (5.30)/30`

`<=> 5(5x+2) - 10(8x-1) =6(4x+2) - 5.30`

`<=> 25x + 10 - 80x + 10 = 24x+12 - 150`

`<=> -55x +20 = 24x-138`

`<=> -55x -24x=-138-20`

`<=>-79x=-158`

`<=> x=2`

Vậy pt có nghiệm `x=2`

`b,` \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)

Ta có : `(x+2)/(x-2) -1/x = 2/(x(x-2))`

`<=> (x(x+2))/(x(x-2)) - (x-2)/(x(x-2))  = 2/(x(x-2))`

`=> x^2 +2x - x +2 = 2`

`<=> x^2 + x =0`

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

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

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

Vậy pt có nghiệm `x=-1`

`c,2x^3 + 6x^2 =x^2 +3x`

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

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

`->` Đề có sai ko ạ ?

`d,` \(\left|x-4\right|+3x=5\) `(1)`

Thường hợp `1` : `x-4 >= 0<=> x >=0` thì phương trình `(1)` thở thành :

`x-4 = 5-3x`

`<=> x+3x=5+4`

`<=> 4x=9`

`<=> x= 9/4 (t//m)`

Trường hợp `2` : `x-4< 0<=> x<0` thì phương trình `(1)` trở thành :

`-(x-4) =5-3x`

`<=> -x +4=5-3x`

`<=> -x+3x=5-4`

`<=> 2x =1`

`<=>x=1/2 ( kt//m)`

Vậy phương trình có nghiệm `x=9/4`

đây là phương trình mà đâu phải bất phương trình đâu