\(|x-6|=-5x+9\)

Xét \(x\ge6\)thì \(pt< =>x-6=-5x+9\)

\(< =>x-6+5x-9=0\)

\(< =>6x-15=0\)

\(< =>x=\frac{15}{6}\)(ktm)

Xét \(x< 6\)thì \(pt< =>x-6=5x-9\)

\(< =>4x-9+6=0\)

\(< =>4x-3=0< =>x=\frac{3}{4}\)(tm)

Vậy ...

\(|x+1|=x^2+x\)

Xét \(x\ge-1\)thì \(pt< =>x+1=x^2+x\)

\(< =>x^2+x-x-1=0\)

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

\(< =>\orbr{\begin{cases}x=1\\x=-1\end{cases}\left(tm\right)}\)

Xét \(x< -1\)thì \(pt< =>-x-1=x^2+x\)

\(< =>x^2+2x+1=0\)

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

\(< =>x=-1\left(ktm\right)\)

Vậy ...

a, \(\left(x-3\right)\left(x^2+x-20\right)\ge0\)

\(\Leftrightarrow\) \(\left(x-3\right)\left(x-4\right)\left(x+5\right)\ge0\)

+) \(x-3=0\Leftrightarrow x=3\); \(x-4=0\Leftrightarrow x=4\); \(x+5=0\Leftrightarrow x=-5\)

+) Lập trục xét dấu f(x) (Bạn tự kẻ trục nha)

\(\Rightarrow\) Bpt có tập nghiệm S = \(\left[-5;3\right]\cup\) [4; \(+\infty\))

b, \(\dfrac{x^2-4x-5}{2x+4}\ge0\)

\(\Leftrightarrow\) \(\dfrac{\left(x-5\right)\left(x+1\right)}{2x+4}\ge0\)

+) \(x-5=0\Leftrightarrow x=5\); \(x+1=0\Leftrightarrow x=-1\); \(2x+4=0\Leftrightarrow x=-2\)

+) Lập trục xét dấu f(x) 

\(\Rightarrow\) Bpt có tập nghiệm S = (-2; -1] \(\cup\) [5; \(+\infty\))

c, \(\dfrac{-1}{x^2-6x+8}\le1\)

\(\Leftrightarrow\) \(\dfrac{\left(x-3\right)^2}{\left(x-4\right)\left(x-2\right)}\ge0\)

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

+) Lập trục xét dấu f(x)

\(\Rightarrow\) Bpt có tập nghiệm S = (\(-\infty\); 2) \(\cup\) (4; \(+\infty\))

Chúc bn học tốt!

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

Vậy \(x\in\left\{\dfrac{2}{3};4\right\}\)

\(b,x^2-6x+9=9\left(x-1\right)^2\\ \Leftrightarrow\left(x-3\right)^2=9\left(x-1\right)^2\\ \Leftrightarrow\left(x-3\right)^2-9\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-3\right)^2-3^2\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-3\right)^2-\left[3\left(x-1\right)\right]^2=0\\ \Leftrightarrow\left(x-3\right)^2-\left(3x-3\right)^2=0\\ \Leftrightarrow\left(x-3+3x-3\right)\left(x-3-3x+3\right)=0\\ \Leftrightarrow-2x\left(4x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-2x=0\\4x-6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{0;\dfrac{3}{2}\right\}\)

1. x² - 6x + 9=(x-3)²

2. 25 +  10x + x²=(x+5)²

3. \(\dfrac{1}{4}a^2+2ab^2+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)

4.\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)

5.x³ + 8y³=(x+8y)(x²-8xy+64y²)

6.8y³ -125=(2y-5)(4y²+10y+25)

7.a⁶-b³=(a²-b)(a⁴+a²b+b²)

8 x² - 10x + 25=(x-2)²

1) \(x^2-6x+9=\left(x-3\right)^2\)

2) \(25+10x+x^2=\left(5+x\right)^2\)

3) \(\dfrac{1}{4}a^2+2ab+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)

4) \(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)

5) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

6) \(8y^3-125=\left(2y-5\right)\left(4y^2+10y+25\right)\)

7) \(a^6-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)

8) \(x^2-10x+25=\left(x-5\right)^2\)

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

\(x=2\)

\(x=0,5\)

\(x=3,125\)

a. 

x = 2 ; 3

b. 

x = 0

c. 

x = 3,125

a, (3 - x) = (3 - x)² 
Suy ra 3-x = 1 hoặc 3-x = 0
--Nếu 3-x = 1
          x = 3-1
          x = 2
--Nếu 3-x = 0
          x = 3 - 0
          x = 3
Vậy x=2; x=3
b, x³ - 0,25x = 0
    x.(x² - 0,25) = 0
Suy ra x = 0   hoặc   x² - 0,25 = 0
                                 x² = 0 + 0,25
                                 x² = 0,25
                                 x² = (0,5)² 
                                 Suy ra x = 0,5
Vậy x = 0; x = 0,5
c, x2 - 10x = -25
    x.(2 - 10) = -25
    x.(-8) = -25
    x = -25 : (-8)
    x = 25/8
Vậy x = 25/8
Chúc bạn học tốt

2 x 2  – x = 3 – 6x

⇔ 2 x 2  – x + 6x – 3 = 0

⇔ (2 x 2  + 6x) – (x + 3) = 0

⇔ 2x(x + 3) – (x + 3) = 0

⇔ (2x – 1)(x + 3) = 0

⇔ 2x – 1 = 0 hoặc x + 3 = 0

2x – 1 = 0 ⇔ x = 1/2

x + 3 = 0 ⇔ x = -3

Vậy phương trình có nghiệm x = 1/2 hoặc x = -3

x 2   +   9   >   6 x   ⇔   ( x   -   3 ) 2  > 0 (đúng với mọi x)