có ai giải cho đâu mà cảm ơn

a, 3x-2=2x-3 <=> 3x-2x=-3+2 <=> x=-1

b, 2x+3=5x+9 <=> 5x-2x=3-9 <=> 3x=-6 <=> x=-2

c, 5-2x=7 <=> 2x=5-7 <=> 2x=-2 <=> x=-1

d, x(x+2)=x(x+3) <=> x^2 + 2x = x^2 + 3x <=> 3x-2x=0 <=> x=0

e, 

a) (x + 6)(3x + 1) + x² - 36 = 0

<=> 3x² + x + 18x + 6 + x² - 36 = 0

<=> 4x² + 19x - 30 = 0

<=> 4x² + 24x - 5x - 30 = 0

<=> 4x(x + 6) - 5(x + 6) = 0

<=> (x + 6)(4x - 5) = 0

<=> x + 6 = 0 hoặc 4x - 5 = 0

<=> x = -6 hoặc x = 5/4

Bài 1 mình đã làm xong rồi, anh em nào giúp mình bài 2 với!

a, \(3x+2\left(x-5\right)=6-\left(5x-1\right)\)

\(\Leftrightarrow3x+2x-10=6-5x+1\)

\(\Leftrightarrow-15\ne0\)Vậy phương trình vô nghiệm 

b, \(x^3-3x^2-x+3=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=3;\pm1\)

Vậy tập nghiệm của phương trình là S = { 1 ; -1 ; 3 }

c, \(\frac{1}{x-3}+\frac{x}{x+3}=\frac{2}{x^2-9}ĐK:x\ne\pm3\)

\(\Leftrightarrow\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)

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

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

Vậy ... 

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

Bài 1:

a) 5(x-3)-4=2(x-1)

\(\Leftrightarrow5x-15-4=2x-2\)

\(\Leftrightarrow5x-19-2x+2=0\)

\(\Leftrightarrow3x-17=0\)

\(\Leftrightarrow3x=17\)

\(\Leftrightarrow x=\frac{17}{3}\)

Vậy: \(x=\frac{17}{3}\)

b) 5-(6-x)=4(3-2x)

\(\Leftrightarrow5-6+x=12-8x\)

\(\Leftrightarrow-1+x-12+8x=0\)

\(\Leftrightarrow-13+9x=0\)

\(\Leftrightarrow9x=13\)

\(\Leftrightarrow x=\frac{13}{9}\)

Vậy: \(x=\frac{13}{9}\)

c) (3x+5)(2x+1)=(6x-2)(x-3)

\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)

\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)

\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)

\(\Leftrightarrow33x-1=0\)

\(\Leftrightarrow33x=1\)

\(\Leftrightarrow x=\frac{1}{33}\)

Vậy: \(x=\frac{1}{33}\)

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

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

\(\Leftrightarrow x^2+6x-4=x^2-6x+8\)

\(\Leftrightarrow x^2+6x-4-x^2+6x-8=0\)

\(\Leftrightarrow12x-12=0\)

\(\Leftrightarrow x=1\)

Vậy:x=1

Bài 2:

a)\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)

\(\Leftrightarrow\frac{x}{3}-\frac{5x}{6}-\frac{5x}{4}-\frac{x}{4}+5=0\)

\(\Leftrightarrow\frac{4x}{12}-\frac{10x}{12}-\frac{15x}{12}-\frac{3x}{12}+\frac{60}{12}=0\)

\(\Leftrightarrow4x-10x-15x-3x+60=0\)

\(\Leftrightarrow-24x+60=0\)

\(\Leftrightarrow-24x=-60\)

\(\Leftrightarrow x=\frac{5}{2}\)

Vậy: \(x=\frac{5}{2}\)

b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow\frac{8x-3}{4}-\frac{3x-2}{2}-\frac{2x-1}{2}-\frac{x+3}{4}=0\)

\(\Leftrightarrow\frac{8x-3}{4}-\frac{2\left(3x-2\right)}{4}-\frac{2\left(2x-1\right)}{4}-\frac{x+3}{4}=0\)

\(\Leftrightarrow8x-3-2\left(3x-2\right)-2\left(2x-1\right)-\left(x+3\right)=0\)

\(\Leftrightarrow8x-3-6x+4-4x+2-x-3=0\)

\(\Leftrightarrow-3x=0\)

\(\Leftrightarrow x=0\)

Vậy: x=0

c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)

\(\Leftrightarrow\frac{15\left(x-1\right)}{30}-\frac{2\left(x+1\right)}{30}-\frac{5\left(2x-13\right)}{30}=0\)

\(\Leftrightarrow15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)=0\)

\(\Leftrightarrow15x-15-2x-2-10x+65=0\)

\(\Leftrightarrow3x+48=0\)

\(\Leftrightarrow3x=-48\)

\(\Leftrightarrow x=-16\)

Vậy: x=-16

d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)

\(\Leftrightarrow\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}-\frac{1-x}{2}+2=0\)

\(\Leftrightarrow\frac{9\left(3-x\right)}{24}+\frac{16\left(5-x\right)}{24}-\frac{12\left(1-x\right)}{24}+\frac{48}{24}=0\)

\(\Leftrightarrow9\left(3-x\right)+16\left(5-x\right)-12\left(1-x\right)+48=0\)

\(\Leftrightarrow27-9x+80-16x-12+12x+48=0\)

\(\Leftrightarrow-13x+143=0\)

\(\Leftrightarrow-13x=-143\)

\(\Leftrightarrow x=11\)

Vậy: x=11

e) \(\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)\)

\(\Leftrightarrow\frac{3\left(5x-2\right)}{4}-2-\frac{7x}{3}+5\left(x-7\right)=0\)

\(\Leftrightarrow\frac{9\left(5x-2\right)}{12}-\frac{24}{12}-\frac{28x}{12}+\frac{60\left(x-7\right)}{12}=0\)

\(\Leftrightarrow9\left(5x-2\right)-24-28x+60\left(x-7\right)=0\)

\(\Leftrightarrow45x-18-24-28x+60x-420=0\)

\(\Leftrightarrow77x-462=0\)

\(\Leftrightarrow77x=462\)

\(\Leftrightarrow x=6\)

Vậy:x=6

Bài 3:

a) \(\left(5x-4\right)\left(4x+6\right)=0\)

\(\Leftrightarrow\left(5x-4\right)\cdot2\cdot\left(2x+3\right)=0\)

Vì \(2\ne0\)

nên \(\left[{}\begin{matrix}5x-4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=\frac{-3}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{4}{5};-\frac{3}{2}\right\}\)

b) \(\left(x-5\right)\left(3-2x\right)\left(3x+4\right)=0\)

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

Vậy: \(x\in\left\{5;\frac{3}{2};\frac{-4}{3}\right\}\)

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

Ta có: \(\left(2x+1\right)\left(x^2+2\right)=0\)(1)

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+2\ge2\ne0\forall x\)(2)

Từ (1) và (2) suy ra:

\(2x+1=0\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy: \(x=\frac{-1}{2}\)

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

\(\Leftrightarrow4\left(2x-1\right)\left(x^2+2x+2\right)=0\)

Ta có: \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\)

Ta lại có \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1\ne0\forall x\)(3)

Ta có: \(4\ne0\)(4)

Từ (3) và (4) suy ra

2x-1=0

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy: \(x=\frac{1}{2}\)

Bài 4:

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

\(\Leftrightarrow2x^2+3x-4x-6=x^2-2x-x+2\)

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

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

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

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

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

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

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

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

Vậy: \(x\in\left\{2;-4\right\}\)

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

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)-\left(x-5\right)\left(4-x\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0\)

\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)

Vì \(3\ne0\)

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

Vậy: \(x\in\left\{0;4\right\}\)

c) \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)

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

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

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

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

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

Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)

d) \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)

\(\Leftrightarrow x^2+4x+4-9\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow x^2+4x+4-9x^2+36x-36=0\)

\(\Leftrightarrow-8x^2+40x-32=0\)

\(\Leftrightarrow-\left(8x^2-40x+32\right)=0\)

\(\Leftrightarrow-8\left(x^2-5x+4\right)=0\)

Vì \(-8\ne0\)

nên \(x^2-5x+4=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

Vậy: \(x\in\left\{1;4\right\}\)

e) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow4\left(4x^2+28x+49\right)-9\left(x^2+6x+9\right)=0\)

\(\Leftrightarrow16x^2+112x+196-9x^2-54x-81=0\)

\(\Leftrightarrow7x^2+58x+115=0\)

\(\Leftrightarrow7x^2+23x+35x+115=0\)

\(\Leftrightarrow x\left(7x+23\right)+5\left(7x+23\right)=0\)

\(\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-23\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-23}{7};-5\right\}\)

Bài 5:

a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)

\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x^2-1\right)=0\)

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

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

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

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

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

Vậy: \(x\in\left\{-\frac{2}{3};-1;\frac{1}{2}\right\}\)

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

\(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)

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

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

\(\Leftrightarrow3x^2-3=0\)

\(\Leftrightarrow3\left(x^2-1\right)=0\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)=0\)

Vì \(3\ne0\)

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

Vậy: \(x\in\left\{1;-1\right\}\)

c) \(x^4+x^3+x+1=0\)

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

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

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

\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2-x+1\right)=0\)(5)

Ta có: \(x^2-x+1=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

Ta lại có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\ne0\forall x\)(6)

Từ (5) và (6) suy ra

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

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy: x=-1

ko khó đâu, chủ yếu nhát làm

Câu 1:

a.5.(x-3)-4=2.(x-1)

⇔5x-15-4=2x-2

⇔ 5x-2x=-2+19

⇔ 3x=17

⇔ x=17/3

b. 5-(6-x)=4.(3-2x)

⇔ x-1=12-8x

⇔ x+8x=12+1

⇔ x=13/9

c.(3x+5).(2x+1)=(6x-2).(x-3)

⇔ 6x² + 3x+10x+5=6x²-18x-2x+6

⇔ (6x²-6x²)+(13x+20x)=6-5

⇔ 33x=1

⇔x=1/33

d.(x+2)²+2.(x-4)=(x-4).(x-2)

⇔x²+4x+4+2x-8=x²-2x-4x+8

⇔(x²-x²)+(6x+6x)=8+8-4

⇔12x=12

⇔ x=1

\(a,9\left(2x+1\right)=4\left(x-5\right)^2\)

\(4x^2-40x+100=18x+9\)

\(4x^2-58x+91=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}}\)

\(c,x^3+3x^2-6x-8=0\)

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

\(Th1:x+4=0\Leftrightarrow x=-4\)

\(Th2:x-2=0\Leftrightarrow x=2\)

\(Th3:x+1=0\Leftrightarrow x=-1\)

\(a,9.\left(2x+1\right)=4.\left(x-5\right)^2\)

\(< =>4x^2-40x+100=18x+9\)

\(< =>4x^2+58x+91=0\)

\(< =>\orbr{\begin{cases}x=\frac{29-3\sqrt{53}}{4}\\x=\frac{29+3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(< =>\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(< =>\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}\)

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