Đại số lớp 8

Mình sửa chút nhé

A=x²+y²-x-y-xy+1

=\(\dfrac{1}{2}\)(2x²+2y²-2x-2y-2xy+2)

=\(\dfrac{1}{2}\)[(x²-2xy+y²)+(x²-2x+1)+(y²-2y+1)]

=\(\dfrac{1}{2}\)[(x-y)²+(x-1)²+(y-1)²]

Min A=0 tại x=y=1

a) \(x^3-7x+6=x^3+3x^2-x^2-3x-2x^2-6x+2x+6\)

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

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

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

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

\(b...x^3-19x+30=0\)

\(=>x^3+5x^2-2x^2-10x-3x^2-15x+6x+30=0\)

=>\(x^2\left(x+5\right)-2x\left(x+5\right)-3x\left(x+5\right)+6\left(x+5\right)=0\)

=>\(\left(x+5\right)\left(x^2-2x-3x+6\right)=0\)

=>\(\left(x+5\right)\left(x-3\right)\left(x-2\right)=0\)

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

Vậy x=-5;2;3

a) đk: \(x\ne\pm3\)

\(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)

\(=\left(-\frac{21}{9-x^2}-\frac{\left(x-4\right)\left(3+x\right)}{9-x^2}-\frac{\left(x-1\right)\left(3-x\right)}{9-x^2}\right):\left(\frac{x+2}{x+3}\right)\)

\(=\frac{-6-3x}{9-x^2}\cdot\frac{x+3}{x+2}=\frac{-3\left(x+2\right)}{9-x^2}\cdot\frac{x+3}{x+2}=\frac{-3}{3-x}\)

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

\(B\left(2\right)=-\frac{3}{3-2}=-3\)

c) \(B=-\frac{3}{5}\Leftrightarrow-\frac{3}{3-x}=-\frac{3}{5}\Leftrightarrow3-x=5\Leftrightarrow x=-2\)

d) \(B< 0\Leftrightarrow-\frac{3}{3-x}< 0\Leftrightarrow3-x>0\Leftrightarrow x< 3\)

a.B=\(\frac{3}{x-3}\)

b.|2x+1|=5

<=> \(\left[\begin{matrix}x=2\Rightarrow B=-3\\x=-3\Rightarrow B=-\frac{1}{2}\end{matrix}\right.\)

c.B=-3/5

\(\frac{3}{x-3}=-\frac{3}{5}\Leftrightarrow x=-3\)

d.\(\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\)(vi 3>0)

<=> x<3

Bài 2

Ta có :

\(x^2+5x+6=\left(x+2\right)\left(x+3\right)\)

\(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

\(x^2+9x+20=\left(x+4\right)\left(x+5\right)\)

Khi đó:

\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}=\dfrac{3}{40}\)

=> \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}=\dfrac{3}{40}\)

=> \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{3}{40}\)

=> \(\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{3}{40}\)

Giải phương trình ta được x = 3

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

\(\Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16=0\)

\(\Leftrightarrow\left(x^2+x+2\right)^2-16=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)

vì \(\left(x^2+x+6\right)>0\) nên\(\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy tập nghiện của pt là:\(S=\left\{1;-2\right\}\)

đặt x^2 + x =y

pt trở thành y^2 +4y -12=0

<=> (y-2)(y+6)=0 <=> y=2 hoặc y=-6

với y=2 => x^2+x=2 <=> x^2 + x-2=0

<=> (x-1)(x+2)=0 <=> x=1 và x=-2

với y=-6=> x^2 + x=-6 <=> x^2 +x+6=0 ( vô nghiệm)

vậy S=-2;1

Ta có :

\(f\left(x\right)=x^3+ax^2+2x+b\)

\(f\left(x\right)=x\left(x^2+x+1\right)+\left(a-1\right)\left(x^2+x+1\right)+\left(2-a\right)x+b-a+1\)\(f\left(x\right)=\left(x+a-1\right)\left(x^2+x+1\right)+\left(2-a\right)x+b-a+1\)

⇒ Để \(f\left(x\right)\) chia hết cho \(g\left(x\right)\)thì

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

⇒\(\left\{{}\begin{matrix}2-a=0\\b-a+1=0\end{matrix}\right.\)

⇒\(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)

\(C=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)...\left(1-\dfrac{1}{210}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\cdot\cdot\dfrac{209}{210}\)

Ta có:

\(n^5-5n^3+4n=n\left(n^4-5n^2+4\right)\)

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

\(=n\left[n^2\left(n^2-1\right)-4\left(n^2-1\right)\right]\)

\(=n\left(n^2-1\right)\left(n^2-4\right)\)

\(=n\left(n-1\right)\left(n+1\right)\left(n-2\right)\left(n+2\right)\)

\(\Rightarrow\)\(\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

Vì \(n-2;n-1;;n;n+1;n+2\) là tích của 5 số nguyên liên tiếp chia hết cho 3;5;8

Mà ƯC\(_{\left(3;5;8\right)}\)=1

\(\Rightarrow\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)\) chia hết cho:

3.5.8=120(đpcm)

A=\(\dfrac{2010x+2680}{x^2+1}\)

A=\(\dfrac{-335x^2-335+335x^2+2010x+3015}{x^2+1}\)

A=\(-335+\dfrac{335\left(x+3\right)^2}{x^2+1}\ge\)\(-335\)

Vậy GTNN A=\(-335\)\(\Leftrightarrow x=-3\)

2680 tại x=0

Min A = -335 khi x=-3