a: A=căn x-1

=>\(\left(\sqrt{x}-1\right)\cdot\dfrac{\sqrt{x}}{3}-\left(\sqrt{x}-1\right)=0\)

=>(căn x-1)(1/3*căn x-1)=0

=>x=1 hoặc x=9

b: \(B=\dfrac{x+2\sqrt{x}+4-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{3}{x-\sqrt{x}+1}\)

c: \(A\cdot B< =\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}\)

=>\(\dfrac{3\left(x-\sqrt{x}\right)}{3\left(x-\sqrt{x}+1\right)}< =\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}\)

=>x-2căn x+1<=0

=>(căn x-1)^2<=0

=>x=1

a:=>(x-3)(2x-5)=0

=>x=3 hoặc x=5/2

b: =>(x-2)(x+2)+(x-2)(2x-3)=0

=>(x-2)(x+2+2x-3)=0

=>(x-2)(3x-1)=0

=>x=1/3 hoặc x=1

c: =>(2x+5-x-2)(2x+5+x+2)=0

=>(x+3)(3x+7)=0

=>x=-7/3 hoặc x=-3

d: =>(x-2)(x-3)=0

=>x=2 hoặc x=3

e: =>(x+3)(x^2-3x+9+x-9)=0

=>(x+3)(x^2-2x)=0

=>\(x\in\left\{0;2;-2\right\}\)

f: =>(10x-5)^2=(2x-4)^2

=>(10x-5-2x+4)(10x-5+2x-4)=0

=>(8x-1)(12x-9)=0

=>x=3/4 hoặc x=1/8

Vì \(RN=NQ,QM=QP\Rightarrow MN\) là đường trung bình tam giác PQR

\(\Rightarrow PQ=2MN=2.22,125=44,25\)

bạn cho mình lời giải chi tiết vs ạ

chụp màn hình câu hỏi kiểu gì vậy bạn

bn có help ko ;-;

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

\(\Rightarrow4x-10=3x+6\)

\(\Rightarrow x=16\)

\(\dfrac{2x+5}{3}=\dfrac{x+2}{2}\)

MTC : 6

Quy đồng mẫu thức :

\(\Rightarrow\) \(\dfrac{2\left(2x+5\right)}{6}=\)\(\dfrac{3\left(x+2\right)}{6}\)

 Suy ra : 2(2x + 5) = 3(x + 2)

          \(\Leftrightarrow\) 4x + 10 = 3x + 6

          \(\Leftrightarrow\) 4x + 10 - 3x - 6 = 0

          \(\Leftrightarrow\) x + 4 = 0

          \(\Leftrightarrow\) x = - 4

       Vậy S = \(\left\{-4\right\}\)

 Chúc bạn học tốt

`(3x+2)/3 <= (x-4)/7`

`<=>7(3x+2) <= 3(x-4)`

`<=>21x+14 <= 3x-12`

`<=> 18x <=-26`

`<=>x <= -13/9`

Vậy `x<=-13/9`.

`(3x+2)/3 <= (x-4)/7`

`<=>7(3x+2) <= 3(x-4)`

`<=> 21x+14<=3x-12`

`<=>18x <= -26`

`<=> x <=-13/9`

thx bn nhìu nha

ĐK: ` x\ne \pm 3`

`(x+1)/(x-3)+(x-1)/(x+3)=(x+6)/(x^2-9)`

`<=>(x+1)(x+3)+(x-1)(x-3)=x+6`

`<=>x^2+4x+3+x^2-4x+3=x+6`

`<=>2x^2+6=x+6`

`<=>2x^2-x=0`

`<=>x(2x-1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy `S={0; 1/2}`.

ĐKXĐ: x ≠ -3, x ≠ 3

\(\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}=\dfrac{x+6}{x^2-9}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)+\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow x^2+4x+3+x^2-4x+3=x+6\)

\(\Leftrightarrow2x^2-x=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)

Vậy...