\(4x^4+9=0\)

\(4x^4=-9\)

mà \(4x^4\ge0\)  với mọi x

=> Vô nghiệm

4x^4+9=0

Suy ra 4x^4=-9

Suy ra 16x^2=-3^2

16x=-3

x=-3/16

x=52/45 hoặc x= -52/45

HT

\(|x|-\dfrac{3}{5}=\dfrac{5}{9}\)

\(|x|=\dfrac{5}{9}+\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{52}{45};\dfrac{-52}{45}\)

a) 18990

b) 5220

a) a=9; b=0

b) x=2; y=0

Mẫu số to quá nên ko nghĩ ra cách giải đẹp mắt:

Dự đoán dấu "=" xảy ra tại \(a=b=c=1\), ta cần c/m: \(A\le\dfrac{3}{16}\)

Do \(\sum\dfrac{a+1}{a^2+1+10a+20}\le\sum\dfrac{a+1}{2a+10a+20}=\sum\dfrac{a+1}{12a+20}\)

Nên ta chỉ cần chứng minh: \(\sum\dfrac{a+1}{3a+5}\le\dfrac{3}{4}\Leftrightarrow\sum\left(\dfrac{3a+3}{3a+5}-1\right)\le\dfrac{9}{4}-3\)

\(\Leftrightarrow\sum\dfrac{1}{3a+5}\ge\dfrac{3}{8}\Leftrightarrow\dfrac{3\left(ab+bc+ca\right)+10\left(a+b+c\right)+25}{\left(3a+5\right)\left(3b+5\right)\left(3c+5\right)}\ge\dfrac{1}{8}\) (quy đồng)

\(\Leftrightarrow\dfrac{4\left(a+b+c\right)+3\left(ab+bc+ca+2\left(a+b+c\right)\right)+25}{27abc+45\left(ab+bc+ca+2\left(a+b+c\right)\right)-15\left(a+b+c\right)+125}\ge\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4\left(a+b+c\right)+52}{27abc-15\left(a+b+c\right)+530}\ge\dfrac{1}{8}\)

\(\Leftrightarrow47\left(a+b+c\right)\ge27abc+114\)

Điều này đúng do:

\(9=2\left(a+b+c\right)+ab+bc+ca\le2\left(a+b+c\right)+\dfrac{1}{3}\left(a+b+c\right)^2\)

\(\Rightarrow\left(a+b+c-3\right)\left(a+b+c+9\right)\ge0\)

\(\Rightarrow a+b+c\ge3\)

Và: \(9=a+b+c+a+b+c+ab+bc+ca\ge9\sqrt[9]{a^4b^4c^4}\)

\(\Rightarrow abc\le1\)

\(\Rightarrow\left\{{}\begin{matrix}47\left(a+b+c\right)\ge141\\27abc+114\le27+114=141\end{matrix}\right.\) (đpcm)

\(8x-75=5x+21\)

\(8x-5x=75+21\)

\(3x=96\)

\(x=32\)

Vậy \(x=32.\)

\(9x+25=-\left(2x-58\right)\)

\(9x+25=-2x+58\)

\(9x+2x=-25+58\)

\(11x=33\)

\(x=3\)

Vậy \(x=3.\)

\(15-\left|2x-1\right|=-8\)

\(\left|2x-1\right|=23\)

\(\Rightarrow\orbr{\begin{cases}2x-1=23\\2x-1=-23\end{cases}\Rightarrow}\orbr{\begin{cases}2x=24\\2x=-22\end{cases}}\Rightarrow\orbr{\begin{cases}x=12\\x=-11\end{cases}}\)

Vậy \(x\in\left\{12;-11\right\}\)

8x-5x=75+21

3x=96

x=32

9x+25=-2x+58

9x+2x=58-25

11x=33

x=3

Phương trình hoành độ giao điểm:

x²=2(m+1)x-m²-9 \(\Leftrightarrow\) x²-2(m+1)x+m²+9=0.

Để d không cắt (P) thì \(\Delta\)'<0 \(\Leftrightarrow\) (m+1)²-(m²+9)<0 \(\Leftrightarrow\) m<4.