Đặt \(f\left(x\right)=x^3-2x^2-6x+a\)

Gọi thương của \(f\left(x\right):\left(x-2\right)\)là \(P\left(x\right)\)

\(\Rightarrow f\left(x\right)=P\left(x\right).\left(x-2\right)\)

Thay \(x=2\)ta có: 

\(8-8-12+a=0\)

\(\Rightarrow a=12\)

Vậy \(a=2\)là giá trị cần tìm

Ukm

It's very hard

l can't do it 

Sorry!

a) \(x^4-x^3-7x^2+x+6=0\)

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

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

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

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

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

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

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

\(\Leftrightarrow2x^2+2xy+y^2+9-6x+\left|y+3\right|=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+x^2-6x+9+\left|y+3\right|=0\)

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

Do \(\left(x+y\right)^2\ge0;\left(x-3\right)^2\ge0;\left|y+3\right|\ge0\forall x;y\)

\(\Rightarrow\hept{\begin{cases}x+y=0\\x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)

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

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

\(\Leftrightarrow\left(2x^2+x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}2x^2+x-2=1\\2x^2+x-2=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x^2+x-3=0\\2x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}=0\\x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2-\frac{25}{16}=0\\\left(x+\frac{1}{4}\right)^2-\frac{9}{16}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2=\frac{25}{16}\\\left(x+\frac{1}{4}\right)^2=\frac{9}{16}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\pm\frac{5}{4}\\x+\frac{1}{4}=\pm\frac{3}{4}\end{cases}}\)

Từ đó tính đc x

d) \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)

\(\Leftrightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)=24\)

\(\Leftrightarrow\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

Đặt \(x^2+5x+5=a\), khi đó pt có dạng:

\(\left(a-1\right)\left(a+1\right)-24=0\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+5x+5=5\\x^2+5x+5=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+5x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+2.x.\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\\left(x+\frac{5}{4}\right)^2=-\frac{15}{4}\left(vn\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

a) x=3/2

b)x=-1

c) x=5

d) x= 5/2

/5x-4/=/x+2/

\(\orbr{\begin{cases}5x-4=x+2\\5x-4=-x+2\end{cases}}suyra\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)

vậy x=3/2 hoặc x=1/2

Ta có : \(\left|5x-4\right|=\left|x+2\right|\)

\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x-x=2+4\\5x+x=-2+4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)

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

\(\Rightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}2x-3x=2+3\\2x+3x=-2+3\end{cases}\Rightarrow}\orbr{\begin{cases}-x=5\\5x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}}\)

c)/2+3x/=/4x-3/

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

\(\Rightarrow\orbr{\begin{cases}3x-4x=-3-2\\3x+4x=3-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=-5\\7x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}}\)

d)/7x+1/-/5x+6|=0

\(\Rightarrow\left|7x+1\right|=\left|5x+6\right|\)

\(\Rightarrow\orbr{\begin{cases}7x+1=5x+6\\7x+1=-\left(5x+6\right)\end{cases}\Rightarrow\orbr{\begin{cases}7x-5x=6-1\\7x+1=-5x-6\end{cases}\Rightarrow}\orbr{\begin{cases}2x=5\\7x+5x=-6-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}}\)

115x ² + x ² -x + 1/4 + 15/4 = (x-1/2) ² +115x ² + 15/4 ≥ 15/4

 ⇒ pt vô nghiệm 

x² + 2x = 0

=> x(x + 2) = 0

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

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

(x - 2) + 3.x² - 6x = 0

=> (x - 2) + 3x² - 3x . 2 = 0

=> (x - 2) + 3x.(x - 2) = 0

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

=> \(\orbr{\begin{cases}1+3x=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=2\end{cases}}\)