a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

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

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

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

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0

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

KL:.....

b) .

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

TH1: x+3=0 <=> x=-3

TH2  x-1=0  <=> x=1

KL:.....

c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 ​\)

KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)

KL:....

a) \(3x^2-4x=0\)

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

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

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

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

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

c) \(9x^2+6x+1=0\)

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

\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)

d) \(x^2-4x=4\)

\(\Leftrightarrow\left(x-2\right)^2=8\)

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

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

1) ĐKXĐ: \(x\ge-2\)

\(pt\Leftrightarrow\dfrac{\sqrt{x+2}}{2}+5\sqrt{x+2}-2\sqrt{x+2}=14\)

\(\Leftrightarrow\dfrac{\sqrt{x+2}+6\sqrt{x+2}}{2}=14\Leftrightarrow7\sqrt{x+2}=28\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

2) ĐKXĐ: \(x\ge0\)

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

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

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

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

4) ĐKXĐ: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x\le-1\end{matrix}\right.\)

\(pt\Leftrightarrow\dfrac{x+1}{2x-1}=4\Leftrightarrow x+1=8x-4\)

\(\Leftrightarrow7x=5\Leftrightarrow x=\dfrac{5}{7}\left(tm\right)\)

5) ĐKXĐ: \(x\ge2\)

\(pt\Leftrightarrow\dfrac{x-2}{3x+1}=36\)

\(\Leftrightarrow x-2=108x+36\Leftrightarrow107x=-38\Leftrightarrow x=-\dfrac{38}{107}\left(ktm\right)\)

Vậy \(S=\varnothing\)

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

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

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

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

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

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

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

a: \(\Leftrightarrow14x=56\)

hay x=4

a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le2\end{matrix}\right.\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x>\dfrac{2\sqrt{14}}{7}\\x< -\dfrac{2\sqrt{14}}{7}\end{matrix}\right.\)

c: ĐKXĐ: \(x=\dfrac{1}{3}\)

d: ĐKXĐ: \(-\dfrac{2}{3}< x\le\sqrt{3}\)

giúp tôi với

1) 2x⁴ - 9x³ + 14x² - 9x + 2 = 0

<=> (2x⁴ - 4x³) - (5x³ - 10x²) + (4x² - 8x) - (x - 2) = 0

<=> 2x³(x - 2) - 5x²(x - 2) + 4x(x - 2) - (x - 2) = 0

<=> (2x³ - 5x² + 4x - 1)(x - 2) = 0

<=> [(2x³ - 2x²) - (3x² - 3x) + (x - 1)](x - 2) = 0

<=> [2x²(x - 1) - 3x(x - 1) + (x - 1)](x - 2) = 0

<=> (2x² - 2x - x + 1)(x - 1)(x - 2) = 0

<=> (2x - 1)(x - 1)²(x - 2) = 0

<=> 2x - 1=0

hoặc x - 1 = 0

hoặc x - 2 = 0

<=> x = 1/2

hoặc x = 1

hoặc x = 2

Vậy S = {1/2; 1; 2}

1) \(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\)

\(\Leftrightarrow2x^3\left(x-1\right)-7x^2\left(x-1\right)+7x\left(x-1\right)-2\left(x-1\right)=0\)

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

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

\(\Leftrightarrow\left(x-1\right)\left[2\left(x-1\right)\left(x^2+x+1\right)-7x\left(x-1\right)\right]=0\)

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

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

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

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

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

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

hoặc \(2x-1=0\)

hoặc \(x-2=0\)

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

Vậy tập nghiệm của phương trình là \(S=\left\{1;\frac{1}{2};2\right\}\)

2) \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow6x^4-3x^3+28x^3-14x^2+26x^2-13x-12x+6=0\)

\(\Leftrightarrow3x^3\left(2x-1\right)+14x^2\left(2x-1\right)+13x\left(2x-1\right)-6\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x^3+14x^2+13x-6\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x^3-x^2+15^2-5x+18x-6\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left[x^2\left(3x-1\right)+5x\left(3x-1\right)+6\left(3x-1\right)\right]=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-1\right)\left(x^2+5x+6\right)=0\)

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

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

hoặc \(3x-1=0\)

hoặc \(x+2=0\)

hoặc \(x+3=0\)

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

Vậy tập nghiệm của phương trình là \(S=\left\{\frac{1}{2};\frac{1}{3};-2;-3\right\}\)

3) Ktra lại đề nhé :D

4) \(x^3-3x^2+3x+8=0\)

\(\Leftrightarrow2x^3+2x^2-5x^2-5x+8x+8=0\)

\(\Leftrightarrow2x^2\left(x+1\right)-5x\left(x+1\right)+8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-5x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\2x^2-5x+8=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\left(TM\right)\\2\left(x-\frac{5}{4}\right)^2+\frac{39}{8}=0\left(L\right)\end{cases}}\)

Vậy x = -1

5) \(x^4+2x^3+x^2=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{0;-1\right\}\)

\(a,\Rightarrow5x+3x^2-3x^2-x+2=6\\ \Rightarrow4x=4\Rightarrow x=1\\ b,\Rightarrow\left(2x+\dfrac{1}{2}-1+2x\right)\left(2x+\dfrac{1}{2}+1-2x\right)=2\\ \Rightarrow\dfrac{3}{2}\left(4x-\dfrac{1}{2}\right)=2\\ \Rightarrow6x-\dfrac{3}{4}=2\\ \Rightarrow6x=\dfrac{11}{4}\\ \Rightarrow x=\dfrac{11}{24}\\ c,\Rightarrow\left(x+3\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)