Bài 4: Bất phương trình bậc nhất một ẩn.

(x+2)² +x(x-1)<2x²+1
x²+4x+4+x²-x<2x²+1
3x+4<1
x< -1

Tách bài riêng ra nhé

tách ra ạ

ĐKXĐ:\(17-x\ge0\Leftrightarrow x\le17\)

\(\left|2x-7\right|=17-x\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=17-x\\2x-7=x-17\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=24\\x=-10\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\left(tm\right)\\x=-10\left(tm\right)\end{matrix}\right.\)

\(ĐK:x\ne0;9\)

\(\dfrac{x+9}{x-9}-\dfrac{1}{x}=\dfrac{9}{x\left(x-9\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+9\right)-\left(x-9\right)}{x\left(x-9\right)}=\dfrac{9}{x\left(x-9\right)}\)

\(\Leftrightarrow x\left(x+9\right)-\left(x-9\right)=9\)

\(\Leftrightarrow x^2+9x-x+9-9=0\)

\(\Leftrightarrow x^2+8x=0\)

\(\Leftrightarrow x\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-8\left(tm\right)\end{matrix}\right.\)

không :))

đk : x khác  0 ;9 

\(x^2+9x-x+9=9\Leftrightarrow x^2+8x=0\Leftrightarrow x=0\left(ktm\right);x=-8\left(tm\right)\)

1: \(\Leftrightarrow x\left(x+1\right)-x+1=2\)

\(\Leftrightarrow x^2+x-x+1=2\)

\(\Leftrightarrow x^2=1\)

=>x=1(loại) hoặc x=-1(nhận)

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

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

=>x(x+5)=0

=>x=0(loại) hoặc x=-5(nhận)

1: 

=>x=1 hoặc x=-1

2: 

=>x(x+5)=0

=>x=0 hoặc x=-5.

1.\(ĐK:x\ne0;1\)

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

\(\Leftrightarrow\dfrac{x\left(x+1\right)-\left(x-1\right)}{x\left(x-1\right)}=\dfrac{2}{x\left(x-1\right)}\)

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

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

\(\Leftrightarrow x^2-1=0\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

2.\(ĐK:x\ne0;6\)

\(\dfrac{x+6}{x-6}-\dfrac{1}{x}=\dfrac{6}{x\left(x-6\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+6\right)-\left(x-6\right)}{x\left(x-6\right)}=\dfrac{6}{x\left(x-6\right)}\)

\(\Leftrightarrow x\left(x+6\right)-\left(x-6\right)=6\)

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

\(\Leftrightarrow x^2+5x=0\)

\(\Leftrightarrow x\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)

1: \(\Leftrightarrow x\left(x+1\right)-x+1=2\)

\(\Leftrightarrow x^2+x-x+1=2\)

\(\Leftrightarrow x^2=1\)

=>x=1(loại) hoặc x=-1(nhận)

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

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

=>x(x+5)=0

=>x=0(loại) hoặc x=-5(nhận)

Lời giải:
Do $a,b,c\in [0;1]$ nên:

$a^2(1-b)\leq 0$

$b^2(1-c)\leq 0$

$c^2(1-a)\leq 0$

Cộng theo vế suy ra: $a^2+b^2+c^2\leq a^2b+b^2c+c^2a$ 

Ta có đpcm.

1: 2x+4<5

=>2x<1

hay x<1/2

2: -3x+2<-5

=>-3x<-7

hay x>7/3

3: \(\dfrac{4x-5}{3}>=\dfrac{7-x}{5}\)

=>20x-25>=21-3x

=>23x>=46

hay x>=2

4: =>15<=2x+3

=>2x>=12

hay x>=6