a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)

\(TH_1:3x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

\(TH_2:-2x-7=0\)

\(\Leftrightarrow x=-\dfrac{7}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)

b) \(2x^3-5x^2+3x=0\)

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

\(\Leftrightarrow2x^2\left(x-1\right)-3x\left(x-1\right)=0\)

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

\(TH_1:x=0\)

\(TH_2:x-1=0\)

\(\Leftrightarrow x=1\)

\(TH_3:2x-3=0\)

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

Vậy pt có tập nghiệm \(S=\left\{0;1;\dfrac{3}{2}\right\}\)

c) \(9x^2-16-x\left(3x+4\right)=0\)

\(\Leftrightarrow\left(9x^2-16\right)-x\left(3x+4\right)=0\)

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

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

\(TH_1:3x+4=0\)

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

\(TH_2:2x-4=0\)

\(\Leftrightarrow x=2\)

Vậy pt có tập nghiệm \(S=\left\{-\dfrac{4}{3};2\right\}\)

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

\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)

\(\Rightarrow20x+16-12=9x-6\)

\(\Leftrightarrow20x-9x=-6-16+12\)

\(\Leftrightarrow11x=-10\)

\(\Leftrightarrow x=-\dfrac{10}{11}\)

Vậy pt có nghiệm duy nhất \(x=-\dfrac{10}{11}\)

a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow3x+1=5x+8\)

\(\Leftrightarrow3x-5x=8-1\)

\(\Leftrightarrow-2x=7\)

\(\Leftrightarrow x=\dfrac{-7}{2}\)

Vậy \(X=\dfrac{-7}{2}\)

b) Ta có: \(2x^3-5x^2+3x=0\)

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

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

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

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

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

Vậy \(x=1\) hoặc \(x=0\) hoặc \(x=\dfrac{3}{2}\)

c) \(9x^2-16-x\left(3x+4\right)=0\)

\(\Leftrightarrow9x^2-16-3x^2-4x=0\)

\(\Leftrightarrow6x^2-4x-16=0\)

\(\Leftrightarrow2\left(3x^2-2x-8\right)=0\)

\(\Leftrightarrow3x^2-6x+4x-8=0\)

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

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

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

Vậy \(x=2\) hoặc \(x=\dfrac{-4}{3}\)

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

\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)

\(\Leftrightarrow20x+16-12=9x-6\)

\(\Leftrightarrow20x+16-12-9x+6=0\)

\(\Leftrightarrow11x+10=0\)

\(\Leftrightarrow x=\dfrac{-10}{11}\)

Vậy \(x=\dfrac{-10}{11}\)

a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\-2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\-2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)

a)

\(\Rightarrow x\left(x-5\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-5=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)

b)

\(\Rightarrow3x\left(x-2\right)-2\left(x-2\right)=0\)

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

\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\3x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{2}{3}\end{array}\right.\)

c)

\(\Rightarrow\left(3x-1\right)\left(5x+x-2\right)=0\)

\(\Rightarrow\left(3x-2\right)^2.2=0\)

\(\Rightarrow3x-2=0\)

\(\Rightarrow x=\frac{2}{3}\)

a/ \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(5x-1-5x\right)=0\Leftrightarrow1-5x=0\Leftrightarrow x=\dfrac{1}{5}\)

Vaayj........

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

Vay......

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

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

Vậy.....

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

\(a)\) \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)

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

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

\(\Leftrightarrow\)\(5x-1=0\)

\(\Leftrightarrow\)\(5x=1\)

\(\Leftrightarrow\)\(x=\frac{1}{5}\)

Vậy \(x=\frac{1}{5}\)

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

Suy ra \(x=0\) hoặc \(x+1=0\) hoặc \(x+2=0\)

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

Vậy \(x=0\) hoặc \(x=-1\) hoặc \(x=-2\)

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

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

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

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

Vậy \(x=\frac{-2}{3}\) hoặc \(x=3\)

Chúc bạn học tốt ~ 

a/ \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)

<=> \(\left(5x-1\right)\left(5x-1-5x\right)=0\)

<=> \(-1\left(5x-1\right)=0\)

<=> \(5x-1=0\)

<=> \(5x=1\)

<=> \(x=\frac{1}{5}\)

b/ \(x\left(x+1\right)\left(x+2\right)=0\)

<=> \(x=0\) hoặc \(\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\)

<=> \(x=0\)hoặc \(\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

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

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

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

<=> \(\orbr{\begin{cases}3x=-2\\x=3\end{cases}}\)

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

a, ĐK: \(x\ge\dfrac{1}{5}\)

\(pt\Leftrightarrow\sqrt{5x^2+x+3}+5x-1-2\sqrt{5x-1}+1+x^2+2x+1=-2\)

\(\Leftrightarrow\sqrt{5x^2+x+3}+\left(\sqrt{5x-1}-1\right)^2+\left(x+1\right)^2=-2\)

\(\Rightarrow\) Phương trình vô nghiệm

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

phương trình đâu vậy

sr nhờ cậu giải l.ại vậy nãy nhầm đề

a) 3x + 2 =0 hoặc 2x - 1 =0 tự giải tiếp

b) 3x -4 =0 ... 2x - 1 =0 ....5x - 2 =0

c) x = 0 hoặc 25 x^2 -9 =0 ( ko biết đúng ko nha)

a) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

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

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

b) Ta có: \(x^2-5x+6=0\)

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

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

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

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

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

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

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

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

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

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết

b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

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

c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)

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

  Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)

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

\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)

a) Ta có: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\Leftrightarrow5x^2+3x-5x-3=3x^2-3x-8x+8\)

\(\Leftrightarrow5x^2-2x-3=3x^2-11x+8\)

\(\Leftrightarrow5x^2-2x-3-3x^2+11x-8=0\)

\(\Leftrightarrow2x^2+9x-11=0\)

\(\Leftrightarrow2x^2+11x-2x-11=0\)

\(\Leftrightarrow x\left(2x+11\right)-\left(2x+11\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+11=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-11\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{11}{2};1\right\}\)

b) Ta có: \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow3x\cdot5\cdot\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\15x-35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\15x=35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)

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

c) Ta có: \(\left(2-3x\right)\left(x-11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow2x-22-3x^2+33x=6x-15x^2-4+10x\)

\(\Leftrightarrow-3x^2+35x-22=-15x^2+16x-4\)

\(\Leftrightarrow-3x^2+35x-22+15x^2-16x+4=0\)

\(\Leftrightarrow12x^2+19x-18=0\)

\(\Leftrightarrow12x^2+27x-8x-18=0\)

\(\Leftrightarrow3x\left(4x+9\right)-2\left(4x+9\right)=0\)

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

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

Vậy: \(S=\left\{-\dfrac{9}{4};\dfrac{2}{3}\right\}\)