b, \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)

\(\Rightarrow x^2-9x+20-x^2+x+2=7\)

\(\Rightarrow-8x+22=7\)

\(\Rightarrow-8x=-15\)

\(\Rightarrow x=\frac{15}{8}\)

c, \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)

\(\Rightarrow3x^2-10x+8=3x^2-27x-3\)

\(\Rightarrow3x^2-10x-3x^2+27x=\left(-3\right)+\left(-8\right)\)

\(\Rightarrow17x=-11\)

\(\Rightarrow x=-\frac{11}{17}\)

d, \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)

\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27+5x-x^3=6x\)

\(\Rightarrow6x=-27\)

\(\Rightarrow x=-\frac{27}{6}\)

\(\Rightarrow x=-\frac{9}{2}\)

e, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)

\(\Rightarrow3x^2-2x-5-3x^2-2x+1=x-4\)

\(\Rightarrow-4=x-4\)

\(\Rightarrow x=0\)

b)    (x - 5)(x - 4) - (x + 1)(x - 2) = 7
<=> x² - 9x + 20 - x² + x + 2 - 7 = 0
<=> 8x - 15 = 0 <=> x = 15/8

c)    (3x - 4)(x - 2) = 3x(x - 9) - 3
<=> 3x² - 10x + 8 = 3x² - 27x - 3
<=> 17x = -11 <=> x = -11/17

d)    (x - 3)(x² + 3x + 9) + x(5 - x²) = 6x
<=> x³ - 27 - x³ + 5x - 6x = 0
<=> x = -27

e)    (3x - 5)(x + 1) - (3x - 1)(x + 1) = x - 4
<=> (x + 1)(3x - 5 - 3x + 1) - x + 4 = 0
<=> -4x - 4 - x + 4 = 0 <=> x = 0

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)

Answer:

\(\left|x-1\right|+\left|x-4\right|=3x\)

Trường hợp 1: \(x>1\)

\(1-x+4-x=3x\)

\(\Rightarrow5-2x=3x\)

\(\Rightarrow5=5x\)

\(\Rightarrow x=1\) (Loại)

Trường hợp 2: \(1\le x\le4\)

\(x-1+4-x=3x\)

\(\Rightarrow3=3x\)

\(\Rightarrow x=1\) (Thoả mãn)

Trường hợp 3: \(x>4\)

\(x-1+x-4=3x\)

\(\Rightarrow2x+5=3x\)

\(\Rightarrow2x-3x=5\)

\(\Rightarrow x=-5\) (Loại)

\(\left|x+1\right|+\left|x+4\right|=3x\)

Có: \(\hept{\begin{cases}\left|x+1\right|\ge0\forall x\inℝ\\\left|x+4\right|\ge0\forall x\inℝ\end{cases}}\)

\(\Rightarrow3x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+1+x+4=3x\)

\(\Rightarrow2x+5=3x\)

\(\Rightarrow x=5\)

\(\left|x\left(x-4\right)\right|=x\)

\(\Rightarrow\orbr{\begin{cases}x\left(x-4\right)=x\\x\left(x-4\right)=-x\end{cases}}\Rightarrow\orbr{\begin{cases}x^2-4x=x\\x^2-4x=-x\end{cases}}\Rightarrow\orbr{\begin{cases}x^2-5x=0\\x^2-3x=0\end{cases}}\)

(Nếu ý này bạn trình bàn trong vở thì làm thành một ngoặc vuông to, trong đó chứa hai ngoặc vuông nhỏ nhé.)

Trường hợp 1: \(\orbr{\begin{cases}x=5\text{(Thoả mãn)}\\x=0\text{(Thoả mãn)}\end{cases}}\)

Trường hợp 2: \(\orbr{\begin{cases}x=3\text{(Thoả mãn)}\\x=0\text{(Loại)}\end{cases}}\)

Vậy \(x=5;x=0;x=3\)

2:

a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8

=>x^2-x-12-x^2+4x+5=8

=>3x-7=8

=>3x=15

=>x=5

b: =>3x^2+3x-2x-2-3x^2-21x=13

=>-20x=15

=>x=-3/4

c: =>x^2-25-x^2-2x=9

=>-2x=25+9=34

=>x=-17

d: =>x^3-1-x^3+3x=1

=>3x-1=1

=>3x=2

=>x=2/3

1) \(\left(x+1\right)\left(x+2\right)-3x\left(x-4\right)=x^2+3x+2-3x^2+12x=-2x^2+15x+2\)

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

\(\Leftrightarrow3x^2-10x+8=3x^2-27x\)

\(\Leftrightarrow17x=-8\Leftrightarrow x=-\dfrac{8}{17}\)

3) \(-3\left(x-4\right)\left(x-2\right)-x^2\left(-3x+18\right)+24x-25\)

\(=-3x^3+6x^2+12x^2-24x+3x^3-18x^2+24x-25=-25\)

a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)

\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)

b) \(\Rightarrow x^2-13x+22=0\)

\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)

c) \(\Rightarrow x^2-3x-10=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

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

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)

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

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

\(\Rightarrow x\left(6x-2-15-6x\right)\)

\(\Rightarrow-16x=0\)

\(\Rightarrow x=0\)

d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)

\(\Rightarrow9x^2-4-4x+4=0\)

\(\Rightarrow9x^2-4x=0\)

\(\Rightarrow x\left(9x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)

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

\(c,2x\left(3x-1\right)-3x\left(5+2x\right)=0\\ \Leftrightarrow6x^2-2x-15x-6x^2=0\\ \Leftrightarrow-17x=0\\ \Leftrightarrow x=0\\ d,\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\\ \Leftrightarrow9x^2-4-4x+4=0\\ \Leftrightarrow9x^2-4x=0\\ \Leftrightarrow x\left(9x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)

a) |x - 1| + |x - 4| = 3x (1)

+) Nếu x < 1 => x - 1 < 0; x - 4 < 0 => |x - 1| = 1 - x; |x - 4| = 4 - x

Khi đó (1) trở thành:

1 - x + 4 - x = 3x

=> 5 - 2x = 3x

=> 5 = 3x + 2x

=> 5 = 5x

=> x = 1 (không thoả mãn điều kiện x < 1)

+) Nếu 1 <= x <= 4 => x - 1 >= 0; x - 4 <= 0

=> |x - 1| = x - 1; |x - 4| = 4 - x

Khi đó (1) trở thành: x - 1 + 4 - x = 3x => 3 = 3x

=> x = 1 (thoả mãn)

b)|x+3| ≥ 0;|x+1| ≥ 0

=>|x+3|+|x+1| ≥ 0

Để |x+3|+|x+1|=3x

thì 3x ≥ 0⇒x ≥ 0

=>x+3 > 0 và x+5 > 0

Ta có: x+3+x+1=3x

=>(x+x)+(3+1)=3x

=>2x+4=3x

=>3x-2x=4

=>x=4

Vậy x=4 thỏa mãn

c) lx(x-4)|=x

⇒ x (x − 4) = ±x 

Nếu x (x − 4) = x

⇒ x² − 4x = x 

⇒ x² − 5x = 0 

⇒ x (x − 5) = 0 

⇒  x = 5

     x = 0

Nếu x (x − 4) = −x

⇒ x² − 4x = −x 

⇒ x² − 3x = 0 

⇒ x (x − 3) = 0 

⇒ x = 0

    x = 3 

Vậy x=0 hoặc x=3 hoặc x=5

mỏi tay quá

câu bạn làm sai đề rùi bạn ơi

câu a,x=1

câu b,x=4

câu c,x=0,3,5

g: \(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4=0\)

\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+36=0\)

\(\Leftrightarrow\left(x+3\right)^2\left(x^2+6x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\sqrt{5}-3\\x=-\sqrt{5}-3\end{matrix}\right.\)