Vì \(\left|-x^2+5x-6\right|\ge0\Rightarrow x^2-5x+6\ge0\)

=> Giải bpt.

ĐKXĐ : x² - 5x + 6 \(\ge0\)

<=> \(\orbr{\begin{cases}x\le2\\x\ge3\end{cases}}\)(1)

Khi đó |-x² + 5x - 6| = x² - 5x + 6 

<=> \(\orbr{\begin{cases}-x^2+5x-6=x^2-5x+6\\-x^2+5x-6=-x^2+5x-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}2\left(x^2-5x+6\right)=0\\\forall x\left(2\right)\end{cases}}\)

Khi 2(x² - 5x + 6) = 0

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

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

Từ (1) ; (2) ; (3) =>  \(\orbr{\begin{cases}x\le2\\x\ge3\end{cases}}\)

Vậy  x \(\le2\text{ hoặc }x\ge3\)là nghiệm phương trình 

(8x - 3)(3x + 2) - (4x + 7)(x + 4) = (4x + 1)(5x - 1) 

<=> (24x² + 7x - 6) - (4x² + 23x + 28) = 20x² + x - 1

<=> 24x² + 7x - 6 - 4x² - 23x - 28 - 20x² - x + 1 = 0

<=> -17x - 33 = 0 

<=> -17x = 33

<=> x = -33/17

Vậy x = -33/17 là giá trị cần tìm 

bỏ hộ mik :1) này nha

Sửa đề (x + 5)² - (x - 5)² - 20x + 2

= x² + 10x + 25 - x² + 10x - 25 - 20x + 2

= 2

=> Biểu thức trên không phụ thuộc vào biến

Sửa đề :

(x + 5)² - (x - 5)² - 20x + 2

= x² + 10x + 26 - x² + 10x - 25 - 20x + 2

= 2

=> Biểu thức trên ko phụ thuộc vào biến.

(10x+9)x-(5x-1)(2x+3)=8

<=>10x²+9x-(10x²+15x-2x-3)=8

<=>10x²+9x-10x²-15x+2x+3=8

<=>-4x+3=8

<=>-4x=5

Bài 1:

a)\(\left(a+12\right)^2=a^2+24a+144\)

b)\(\left(3a+\frac{1}{3}\right)^2=9a^2+2a+\frac{1}{9}\)

c)\(\left(5a^2+6\right)^2=25a^4+60a^2+36\)

d)\(\left(\frac{1}{2}+4b\right)^2=\frac{1}{4}+4b+16b^2\)

e)\(\left(a^m+b^n\right)^2=a^{2m}+2a^mb^n+b^{2n}\)

Bài 2:

a)\(\left(x-0,3\right)^2=\left(x-\frac{3}{10}\right)^2=x^2-\frac{3}{5}x+\frac{9}{100}\)

b)\(\left(6x-3y\right)^2=36x^2-36xy+9y^2\)

c)\(\left(5-2xy\right)^2=25-20xy+4x^2y^2\)

d)\(\left(x^4-1\right)^2=x^8-2x^4+1\)

e)\(\left(x^5-y^3\right)^2=x^{10}-2x^5y^3+y^6\)

a) \(A=3x^2+x-1=3\left(x^2+\frac{x}{3}+\frac{1}{36}\right)-\frac{13}{12}=3\left(x+\frac{1}{6}\right)^2-\frac{13}{12}\ge-\frac{13}{12}\forall x\)

Dấu"=" xảy ra \(\Leftrightarrow x+\frac{1}{6}=0\)\(\Leftrightarrow x=-\frac{1}{6}\)

Vậy \(MinA=-\frac{13}{12}\Leftrightarrow x=-\frac{1}{6}\)

b)\(B=t^2-6t=\left(t^2-6t+9\right)-9=\left(t-3\right)^2-9\ge-9\forall t\)

Dấu "=" xảy ra \(\Leftrightarrow t-3=0\)\(\Leftrightarrow t=3\)

Vậy \(MinB=-9\Leftrightarrow t=3\)

c)\(C=x^2+\frac{3}{2}y^2-2x-4y+4\)

\(=\left(x^2-2x+1\right)+\frac{3}{2}\left(y^2-\frac{8}{3}y+\frac{16}{9}\right)+\frac{1}{3}\)

\(=\left(x-1\right)^2+\frac{3}{2}\left(y-\frac{4}{3}\right)^2+\frac{1}{3}\ge\frac{1}{3}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-\frac{4}{3}=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{4}{3}\end{cases}}\)

Vậy \(MinC=\frac{1}{3}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{4}{3}\end{cases}}\)

d)\(D=2x^2+y^2-2xy+4x+2024\)

\(=\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)+2020\)

\(=\left(x-y\right)^2+\left(x+2\right)^2+2020\ge2020\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y=0\\x+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y\\x=-2\end{cases}}\)\(\Leftrightarrow x=y=-2\)

Vậy \(MinD=2020\Leftrightarrow x=y=-2\)

\(x^2+y^2+\frac{8xy}{x+y}=16\)

\(\Leftrightarrow\left(x^2+y^2\right)\left(x+y\right)+8xy-16\left(x+y\right)=0\)

\(\Leftrightarrow\left(x^2+y^2\right)\left(x+y-4\right)+4x^2+4y^2+8xy-16\left(x+y\right)=0\)

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

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

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

\(\Leftrightarrow y=4-x\).

\(Q=x^2-2x+4y+100=x^2-2x+4\left(4-x\right)+100\)

\(=x^2-6x+116=\left(x-3\right)^2+107\ge107\)

Dấu \(=\)khi \(x=3\Rightarrow y=1\).

a) \(x^2-6x-17=\left(x^2-6x+9\right)-26=\left(x-3\right)^3-26\ge-26\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\)

                        \(\Leftrightarrow x-3=0\)

                        \(\Leftrightarrow x=3\)

b)\(x^2-10x=\left(x^2-10x+25\right)-25=\left(x-5\right)^2-25\ge-25\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-5\right)^2=0\)

                        \(\Leftrightarrow x-5=0\)

                        \(\Leftrightarrow x=5\)

c)\(3x^2-12x+5=3\left(x^2-4x+4\right)-7=3\left(x-2\right)^2-7\ge-7\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\)

                        \(\Leftrightarrow x-2=0\)

                        \(\Leftrightarrow x=2\)

d)\(2x^2-x+1=2\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{7}{8}=2\left(x-\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-\frac{1}{4}\right)^2=0\)

                        \(\Leftrightarrow x-\frac{1}{4}=0\)

                        \(\Leftrightarrow x=\frac{1}{4}\)

e)\(x^2+y^2-8x+4y+27=\left(x^2-8x+16\right)+\left(y^2+4y+4\right)+7=\left(x-4\right)^2+\left(y+2\right)^2+7\ge7\forall x\)Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-4\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-4=0\\y+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=4\\y=-2\end{cases}}\)

f)\(x\left(x-6\right)=x^2-6x=\left(x^2-6x+9\right)-9=\left(x-3\right)^2-9\ge-9\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\)

                        \(\Leftrightarrow x-3=0\)

                        \(\Leftrightarrow x=3\)

h)\(\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)\left(x-5\right)\left(x-2\right)\left(x-5\right)=\left[\left(x-2\right)\left(x-5\right)\right]^2\ge0\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-2\right)\left(x-5\right)=0\)

                        \(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}}\)

                        \(\Leftrightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}\)