a.(x+2)²-x(x+2)=0

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

\(\Leftrightarrow\)(x+2)*2=0

\(\Leftrightarrow\)x+2=0

\(\Leftrightarrow\)x=-2

vay s={-2}

b.\(\frac{2x+7}{3}\)-\(\frac{x-2}{4}\)=2

\(\Leftrightarrow\)\(\frac{4\left(2x+7\right)}{12}\)+\(\frac{-3\left(x-2\right)}{12}\)=\(\frac{24}{12}\)

\(\Leftrightarrow\)8x+28-3x+6=24

\(\Leftrightarrow\)5x=-10

\(\Leftrightarrow\)x=-2

vay s={-2}

c.|x+5|=3x+1

neu x+5\(\ge\)0 thi |x+5|=x+5

\(\Leftrightarrow\)x\(\ge\)-5

ta co phuong trinh

x+5=3x+1

\(\Leftrightarrow\)-2x=-4

\(\Leftrightarrow\)x=2( thoa man dieu kien x\(\ge\)-5)

neu x+5<0 thi |x+5|=5-x

\(\Leftrightarrow\)x<-5

ta co phuong trinh

5-x=3x+1

\(\Leftrightarrow\)-4x=-4

\(\Leftrightarrow\)x=1 (k thoa man dieu kien x<5)

vay s={2}

chuc bn hoc tot

a, -2

b, -2

c, 2

(x+2)(x+2-x)=0

(x+2). 2 = 0

<=> x +2 = 0

<=> x = -2

b , \(\frac{2x+7}{3}-\frac{x-2}{4}=2\)

\(\frac{4\left(2x+7\right)}{12}-\frac{3\left(x-2\right)}{12}=\frac{24}{12}\)

=> 8x + 28 - 3x + 6 = 24

<=> 5x = -10

=> x = -2

c , \(|x+5|=3\)x +1

\(|x+5|=x+5\) <=> x + 5 \(\ge0\)

<=> x \(\ge\) -5

\(|x+5|=3\)

\(x+5=3\)

x =-2 (tm)

+ \(|x+5|=-3\Leftrightarrow x+5< 0=>x< -5\)

\(|x+5|=-3\)

Lời giải:

\(\frac{2x-2\sqrt{x}+2}{x-\sqrt{x}}=\frac{2(x-\sqrt{x})+2}{x-\sqrt{x}}=\frac{2(x-\sqrt{x})+2}{x-\sqrt{x}}=2+\frac{2}{x-\sqrt{x}}\)

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

\(x^2+2y^2-2xy+4y+3< 0\)

\(\Rightarrow x^2-2xy+y^2+y^2+4y+4-1< 0\)  

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)-1< 0\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Mà: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\) 

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1\ge-1\forall x,y\)

Mặt khác: \(\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Dấu "=" xảy ra:

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

\(\Rightarrow x=y=-2\)

Vậy: .... 

TH1:  \(m=-1\) thỏa mãn (dễ dàng kiểm tra các giá trị \(f\left(-1\right)>0\) ; \(f\left(0\right)< 0\) ; \(f\left(3\right)>0\) nên pt có ít nhất 2 nghiệm thuộc (-1;0) và (0;3)

TH2: \(m>-1\):

\(\lim\limits_{x\rightarrow+\infty}f\left(x\right)=\lim\limits_{x\rightarrow+\infty}x^4\left[m\left(1-\dfrac{2}{x}\right)^2\left(1+\dfrac{9}{x}\right)+1-\dfrac{32}{x^4}\right]=+\infty.\left(m+1\right)=+\infty>0\)

\(\Rightarrow\) Luôn tồn tại 1 giá trị \(x=a\) đủ lớn sao cho \(f\left(a\right)>0\)

\(f\left(0\right)=-32< 0\Rightarrow f\left(a\right).f\left(0\right)< 0\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm dương

\(f\left(-9\right)=9^4-32>0\Rightarrow f\left(-9\right).f\left(0\right)< 0\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm âm thuộc \(\left(-9;0\right)\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 2 nghiệm

TH3: \(m< -1\) tương tự ta có: \(\lim\limits_{x\rightarrow+\infty}f\left(x\right)=\lim\limits_{x\rightarrow-\infty}=+\infty.\left(m+1\right)=-\infty\)

\(\Rightarrow\) Luôn tồn tại 1 giá trị \(x=a>0\) đủ lớn và \(x=b< 0\) đủ nhỏ sao cho \(\left\{{}\begin{matrix}f\left(a\right)< 0\\f\left(b\right)< 0\end{matrix}\right.\)

Lại có \(f\left(-9\right)=9^4-32>0\) \(\Rightarrow\left\{{}\begin{matrix}f\left(-9\right).f\left(a\right)< 0\\f\left(-9\right).f\left(b\right)< 0\end{matrix}\right.\)

\(\Rightarrow\) Pt luôn có ít nhất 2 nghiệm thuộc  \(\left(-\infty;-9\right)\) và \(\left(-9;+\infty\right)\)

Vậy pt luôn có ít nhất 2 nghiệm với mọi m

a)2x+7=3x+10

\(\Rightarrow7-10=3x-2x\)

\(\Rightarrow-3=x\)

Vậy x=-3

b)Bạn tự làm nha

c)Bạn làm tương tự câu d nha

d)+)Ta có :\(x+1⋮x+1\left(1\right)\)

+)Theo bài ta có:\(x-3⋮x+1\left(2\right)\)

+)Từ (1) và (2)

\(\Rightarrow\left(x+1\right)-\left(x-3\right)⋮x+1\)

\(\Rightarrow x+1-x+3⋮x+1\)

\(\Rightarrow4⋮x+1\)

\(\Rightarrow x+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Rightarrow x\in\left\{-2;0;-3;1;-5;3\right\}\)

Vậy \(x\in\left\{-2;0;-3;1;-5;3\right\}\)

Chúc bn học tốt

a/

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

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

\(\Leftrightarrow6-6x=0\)

=> x=1

b, \(4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4\)

\(\Leftrightarrow12x+9+48=-4x+4\Leftrightarrow16x=-53\Leftrightarrow x=-\dfrac{53}{16}\)

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

\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\Leftrightarrow x=3;x=\dfrac{1}{4}\)

d, \(x^2\left(x+4\right)-9\left(x+4\right)=0\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\Leftrightarrow x=-3;3;-4\)

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

\(3\le x^3\le27\Leftrightarrow x\in\left\{2;3\right\}\\ B=\left\{2;3\right\}\)

\(\Leftrightarrow A\cup B=\left\{-2;0;1;2;3\right\}\)