a) Đặt \(2x^2-3x-1=a\)

Bt \(\Leftrightarrow a^2-3\left(a-4\right)-16=0\)

\(\Leftrightarrow a^2-3a+12-16=0\)

\(\Leftrightarrow\left(a-4\right)\left(a+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-1=4\\2x^2-3x-1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-5=0\\2x^2-3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)\left(2x-5\right)=0\\x\left(2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=-1\\x=2.5\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=1.5\end{matrix}\right.\end{matrix}\right.\)

thay x=-5/4 vào=>f(-5/4)=0
chia x-2 dư 39 =>f(2)=39
đc hệ pt bậc nhất 2 ẩn => tìm đc a và b

a) Áp dụng định lí Py-ta-go vào \(\Delta ABC\) vuông tại A có :

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow BC^2=6^2+8^2=100\)

\(\Leftrightarrow BC=10\)(cm)(vì BC > 0)

\(x\in\left\{-4;-3\right\}\)

\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3\right)-\left(2x^2+1\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+12\right)=0\)

\(\Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2+1=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2=-1\\3x=-9\end{matrix}\right.\Leftrightarrow}x=-3}\)

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

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\) (1)

Đặt \(x^2+5x+4=a\)

\(M=a\left(a+2\right)+1\)

\(=a^2+2a+1=\left(a+1\right)^2=\left(x^2+5x+4\right)^2\)

Mà x nguyên \(\Rightarrow\) M là số chính phương của 1 số nguyên

Bài 1.

b) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

\(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

\(\Leftrightarrow300-x=0\) (vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\))

\(\Leftrightarrow x=300\)

Vậy ....