Câu hỏi của Thị Kim Vĩnh Bùi - Toán lớp 8 - Học toán với OnlineMath

Thay các giá trị a, b, c, d vào M nhận đc giá trị M = 0

Áp dụng BĐT Cô si ta có:

\(x^3+8y^3+1\ge3\sqrt[3]{x^3\cdot8y^3\cdot1}=6xy\)

\(\Rightarrow x^3+8y^3+1-6xy\ge0\)

Dấu "=" xảy ra tại \(x=2y=1\Rightarrow x=1;y=\frac{1}{2}\)

Khi đó:

\(A=x^{2018}+\left(y-\frac{1}{2}\right)^{2019}=1^{2018}+0^{2019}=1\)

\(\left(x-1\right)\left(x+2\right)-x-2=0\)

\(\left(x-1\right)\left(x+2\right)-\left(x+2\right)=0\)

\(\left(x-1-1\right)\left(x+2\right)=0\)

\(\left(x-2\right)\left(x+2\right)=0\)

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

\(\left(x-1\right)\left(x+2\right)-x-2=0\)

\(\left(x-1\right)\left(x+2\right)-\left(x+2\right)=0\)

\(\left(x+2\right)\left(x-1-1\right)=0\)

\(\left(x+2\right)\left(x-2\right)=0\)

\(\Rightarrow x+2=0\)hoặc\(x-2=0\)

\(x=-2\)hoặc\(x=2\)

a) \(8x^3-18x^2+x+6\)

\(=8x^3-16x^2-2x^2+4x-3x+6\)

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

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

\(=\left(x-2\right)\left(8x^2-6x+4x-3\right)\)

\(=\left(x-2\right)\left[2x\left(4x-3\right)+\left(4x-3\right)\right]\)

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

=> g(x) có 3 nghiệm là

x-2=0 <=> x=2

2x+1=0 <=> x=-1/2

4x-3=0 <=> x=3/4

vậy đa thức g(x) có nghiệm là x={2;-1/2;3/4}

b) tự làm đi (mk ko bt làm)

a) \(\frac{1}{x^2-x+1}+1-\frac{x^2+2}{x^3+1}\)

+) Đkxđ: \(\hept{\begin{cases}x^2-x+1\ne0\\x^3+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\ne0\\x^3\ne-1\end{cases}\Leftrightarrow}\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\left(lđ\right)\\x\ne-1\end{cases}}}\)

+) \(A=\frac{1}{x^2-x+1}+1-\frac{x^2+2}{x^3+1}\)

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

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

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

P/s: ko chắc

Huhu luoi qua

a) \(\frac{1}{x^2-x+1}+1-\frac{x^2+2}{x^3+1}\)

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

\(=\frac{x^5-2x^4+3x^3-2x^2+x}{x^5-x^4+x^3+x^2-x+1}\)

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

\(=\frac{x}{x+1}\)

b) \(\frac{7}{x}-\frac{x}{x+6}+\frac{36}{x^2+6x}\)

\(=\frac{-x^2+7x+78}{x^2+6x}\)

\(=\frac{\left(-x-6\right)\left(x-13\right)}{x\left(x+6\right)}\)

\(=\frac{-x+13}{x}\)

Bài làm

a) \(\frac{4x-5}{8xy}+\frac{5-y}{8xy}=\frac{4x-5+5-y}{8xy}=\frac{4x-y}{8xy}\)

b) \(\frac{4x^2}{x-2}+\frac{3}{x-2}+\frac{19}{2-x}=\frac{4x^2}{x-2}+\frac{3}{x-2}-\frac{19}{x-2}=\frac{4x^2+3-19}{x-2}=\frac{4x^2-16}{x-2}=\frac{2\left(x-2\right)\left(2x+4\right)}{x-2}=2\left(2x+4\right)\)

c) \(\frac{2x^3+5}{x^2-x+1}-\frac{x^3+4}{x^2-x+1}=\frac{2x^3+5-x^3-4}{x^2-x+1}=\frac{2x^2-x^3+1}{x^2-x+1}\)

d) \(\frac{6}{5x-20}-\frac{x-5}{x^2-8x+16}=\frac{6}{5\left(x-4\right)}-\frac{x-5}{\left(x-4\right)^2}=\frac{6\left(x-4\right)}{5\left(x-4\right)^2}-\frac{\left(x-5\right)5}{5\left(x-4\right)^2}=\frac{6x-4-5x+25}{5\left(x-4\right)^2}=\frac{x+21}{5\left(x-4\right)^2}\)

# Học tốt #