Ta có: \(x^4+2x^3+8x+16\)

\(=x^3\left(x+2\right)+8\left(x+2\right)\)

\(=\left(x+2\right)\left(x^3+8\right)=\left(x+2\right)^2\cdot\left(x^2-2x+4\right)\)

Ta có: \(x^4-2x^3+8x^2-8x+16\)

\(=x^4-2x^3+4x^2+4x^2-8x+16\)

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

Ta có: \(E=\frac{x^4+2x^3+8x+16}{x^4-2x^3+8x^2-8x+16}\)

\(=\frac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(x^2+4\right)\cdot\left(x^2-2x+4\right)}=\frac{\left(x+2\right)^2}{x^2+4}=\frac{x^2+4x+4}{x^2+4}=1+\frac{4x}{x^2+4}\)

Đặt \(A=\frac{4x}{x^2+4}\)

=>\(A\left(x^2+4\right)=4x\)

=>\(x^2\cdot A-4x+4A=0\) (1)

\(\Delta=\left(-4\right)^2-4\cdot A\cdot4A=16-16A^2\)

Để (1) có nghiệm thì Δ>=0

=>\(16-16\cdot A^2\ge0\)

=>\(A^2\le1\)

=>-1<=A<=1

=>0<=A+1<=2

=>0<=E<=2

=>GTNN của E là E=0 khi A=-1

(1) khi đó sẽ trở thành:

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

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

=>\(\left(x-2\right)^2=0\)

=>x-2=0

=>x=2(nhận)

chào các bạn,có 2 tốt bụng thì tk mik nhé,cần lắm những người như thế

a) |9-7x| = 5x-3

*  9-7x=5x-3    <=>x=1

*  9-7x=-(5x-3) <=> x=3

Tử \(x^4+2x^3+8x+16\)

\(=x^4-2x^3+4x^2+4x^3-8x^2+16x+4x^2-8x+16\)

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

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

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

Mẫu \(x^4-2x^3+8x^2-8x+16\)

\(=x^4-2x^3+4x^2+4x^2-8x+16\)

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

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

Thay tử và mẫu vào ta có:\(\frac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(x^2+4\right)\left(x^2-2x+4\right)}=\frac{\left(x+2\right)^2}{x^2+4}\ge0\)

Dấu "=" khi \(\left(x+2\right)^2=0\Leftrightarrow x=-2\)

Vậy Min=0 khi x=-2

a: =>|7x-9|=5x-3

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;3\right\}\)

b: \(\Leftrightarrow\left|4x+1\right|=8x-x-2=7x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{7}\\\left(7x-2-4x-1\right)\left(7x-2+4x+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{7}\\\left(3x-3\right)\left(11x-1\right)=0\end{matrix}\right.\Leftrightarrow x=1\)

c: |17x-5|=|17x+5|

=>17x-5=17x+5 hoặc 17x+5=5-17x

=>x=0

mình cũng đang bí bài này ai giúp với

1.\(\left|9-7x\right|=5x-3\)

\(\Leftrightarrow\orbr{\begin{cases}9-7x=5x-3\\9-7x=-5x-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-7x-5x=-9-3\\-7x+5x=-9-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-12x=-12\\-2x=-12\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-12:\left(-12\right)\\x=-12:\left(-2\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)

2.\(8x-\left|4x+1\right|=x+2\)

\(\Rightarrow\left|4x+1\right|=8x-x+2\)

\(\Rightarrow\left|4x+1\right|=7x+2\)

\(\Leftrightarrow\orbr{\begin{cases}4x+1=7x+2\\4x+1=-7x+2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x-7x=2-1\\4x+7x=2-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-3x=1\\11x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1:\left(-3\right)\\x=1:11\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{1}{11}\end{cases}}\)

f(0)=-4/10

a/b=-4/10=-2/5

f(1)=-6/26=-3/13=(a+1)/(b+1)

5a=-2b

a/-2=b/5=(a+b)/3

13a+13=-3b-3

15a=-6b

26a=-6b-6

11a=-6

a+b=-3/2.a=3/2.6/11=9/11

a+b=9/11