`|x-2|=2x-3(x>=3/2)`

`<=>` \(\left[ \begin{array}{l}x-2=2x-3\\x-2=3-2x\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=1(l)\\3x=5\end{array} \right.\) 

`<=>x=5/3(Tm(`

`2)A=-x^2+2x+9`

`=-(x^2-2x)+9`

`=-(x^2-2x+1)+1+9`

`=-(x-1)^2+10<=10`

Dấu "=" xảy ra khi `x=1.`

1,

* \(|x-2|=x-2< =>x\ge2\)

\(=>x-2=2x-3< =>x=1\left(ktm\right)\)

*\(\left|x-2\right|=2-x< =>x< 2\)

\(=>2-x=2x-3< =>x=\dfrac{5}{3}\left(tm\right)\)

vậy x=5/3

2, \(A=-x^2+2x+9=-\left(x^2-2x-9\right)=-\left(x^2-2x+1-10\right)\)

\(=-\left[\left(x-1\right)^2-10\right]=-\left(x-1\right)^2+10\le10\)

dấu"=" xảy ra<=>x=1

Bài 1: 

Ta có: \(\left|x-2\right|=2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2x-3\left(x\ge2\right)\\2-x=2x-3\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2x=-3+2\\-x-2x=-3-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=-1\\-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(lọai\right)\\x=\dfrac{5}{3}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)

a) 10-x-5=-5-7-11

=> 5 - x = -23

=> x = 28
b) |x| -3=0

=> |x| = 3

=> x = 3 hoặc x -3
c) ( 7-|x| ) .(2x-4)=0

=> 7 - |x| = 0 hoặc 2x - 4 = 0

=> |x| = 7 hoặc 2x = 4

=> x = 7 hoặc x = - 7 hoặc x = 2
c)2+3x=-15-19 

=> 2 + 3x = -34

=> 3x = 36

=> x = 12 
 

\(a,10-x-5=-5-7-11\)

\(10-x-5=-23\)

\(10-x=-18\)

\(x=28\)

\(b,|x|-3=0\)

\(|x|=3\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}=2x+1\Rightarrow\left|x-3\right|=2x+1\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

a, (x+3)*(y+2)=1

=> x+3 và y+2 là ước của 1

Ta có bảng sau:

Vậy...

i don't know

thế thì trả lời làm gì hả.

k minh minh giai cho

a, 2x + 12= 3(x - 7)

=> 2x + 12 = 3x + 21

=> 12 - 21 = 3x - 2x

=> -9 = x

vậy x = -9

b,-2x-(-17)=15

=> -2x + 17 = 15

=> -2x = 32

=> x = -16

Bài 2

a, A=(-a-b-c)-(-a-b-c)

= -a - b - c + a + b + c 

= 0

b, thay vào thì nó vẫn = 0 thôi

<=> x(y+2)=y+5

=> x=\(\frac{y+5}{y+2}=\frac{y+2+3}{y+2}=1+\frac{3}{y+2}\)

=> để x nguyên thì 3 phải chia hết cho y+2.

=> +/ y+2=1 => y=-1 => x=1+3=4

     +/ y+2=3 => y=1 => x=1+1=2

xy+2x-y=5

=> x(y+2) - y -2 = 5-2

=> x(y+2) - (y+2) = 5 - 2

=> (y+2)(x-1) = 3

do x, y thuộc Z => y+2 và x-1 thuộc Z

=> y+2 và x-1 thuộc Ư(3)={1,-1,-3,3}

chú ý: e là thuộc nhé

Vậy (x,y) e {(-2;-3);(4;-1);(0;-5);(2;1)}

chúc bạn học giỏi

chắc chắn 100% đó

tk nha

c) \(\left(34-2x\right)\left(2x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6-0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)

d) \(\left(2019-x\right)\left(3x-12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)

e) \(57\left(9x-27\right)=0\)

\(\Rightarrow9x-27=0\)

\(\Rightarrow9\left(x-3\right)=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

f) \(25+\left(15-x\right)=30\)

\(\Rightarrow25+15-x=30\)

\(\Rightarrow40-x=30\)

\(\Rightarrow x=40-30\)

\(\Rightarrow x=10\)

g) \(43-\left(24-x\right)=20\)

\(\Rightarrow43-24+x=20\)

\(\Rightarrow19+x=20\)

\(\Rightarrow x=20-19\)

\(\Rightarrow x=1\)

h) \(2\left(x-5\right)-17=25\)

\(\Rightarrow2\left(x-5\right)=17+25\)

\(\Rightarrow x-5=21\)

\(\Rightarrow x=21+5\)

\(\Rightarrow x=26\)

i) \(3\left(x+7\right)-15=27\)

\(\Rightarrow3\left(x+7\right)=27+15\)

\(\Rightarrow x+7=14\)

\(\Rightarrow x=14-7\)

\(\Rightarrow x=7\)

j) \(15+4\left(x-2\right)=95\)

\(\Rightarrow4\left(x-2\right)=95-15\)

\(\Rightarrow4\left(x-2\right)=80\)

\(\Rightarrow x-2=20\)

\(\Rightarrow x=20+2\)

\(\Rightarrow x=22\)

k) \(20-\left(x+14\right)=5\)

\(\Rightarrow x+14=20-5\)

\(\Rightarrow x+14=15\)

\(\Rightarrow x=15-14\)

\(\Rightarrow x=1\)

l) \(14+3\left(5-x\right)=27\)

\(\Rightarrow3\left(5-x\right)=27-14\)

\(\Rightarrow3\left(5-x\right)=13\)

\(\Rightarrow5-x=\dfrac{13}{3}\)

\(\Rightarrow x=5-\dfrac{13}{3}\)

\(\Rightarrow x=\dfrac{2}{3}\)