1200-31/40=47969/40

câu 2 =45/56+25/56=(45+25)/56=70/56=5/4

Đặt : \(\dfrac{x}{5}=\dfrac{y}{3}=k\)

`=>x=5k,y=3k`

Ta có : \(x^2-y^2=4=>\left(5k\right)^2-\left(3k\right)^2=4\\ =>25k^2-9k^2=4\\ =>16k^2=4\\ =>k^2=\dfrac{1}{4}\\ =>k=\pm\dfrac{1}{2}\)

\(=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

a,sửa đề : đk x khác -2;  2 

 \(x^2+x-2+5x-10=12+x^2-4\)

\(\Leftrightarrow6x-20=0\Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)

b, \(3x-12+5+5x=105\Leftrightarrow8x=112\Leftrightarrow x=14\)

c, \(3x^2+14x-49=-\left(x^2+2x-15\right)\)

\(\Leftrightarrow4x^2+16x-34=0\Leftrightarrow x=\dfrac{-4\pm5\sqrt{2}}{2}\)

a. ko hỉu đề lắm :v

b.\(\dfrac{x-4}{5}+\dfrac{1+x}{3}=7\)

\(\Leftrightarrow\dfrac{3\left(x-4\right)+5\left(1+x\right)}{15}=\dfrac{105}{15}\)

\(\Leftrightarrow3\left(x-4\right)+5\left(1+x\right)=105\)

\(\Leftrightarrow3x-12+5+5x-105=0\)

\(\Leftrightarrow8x-112=0\)

\(\Leftrightarrow8x=112\)

\(\Leftrightarrow x=14\)

c.\(\left(3x-7\right)\left(x+7\right)=\left(5+x\right)\left(3-x\right)\)

\(\Leftrightarrow3x^2+21x-7x-49=15-5x+3x-x^2\)

\(\Leftrightarrow4x^2+16x-64=0\)

Nghiệm xấu lắm bạn

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

a \(\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)

b \(\dfrac{22}{77}-\dfrac{14}{77}=\dfrac{8}{77}\)

c \(\dfrac{11}{13}\times\dfrac{26}{31}=11\times\dfrac{2}{31}=\dfrac{22}{31}\)

d \(\dfrac{1}{2}\times3\times\dfrac{2}{5}=\dfrac{3}{5}\)

a, 2/5 + 11/15 = 6/15 + 11/15 = 17/15

b,2/7 - 2/11 = 22/77 - 14/77 = 8/77

c, 11/13 x 26/31= 11 x 2/31 =22/31

d, 1/2 : 1/3 x 2/5

=  ( 1/2 x 2/5 ) : 1/3

=     1/5        : 1/3

=         3/5

Lời giải:
ĐKXĐ: $x\geq 5$

$2x^2-8x-6=2\sqrt{x-5}\leq (x-5)+1$ theo BĐT Cô-si

$\Leftrightarrow 2x^2-9x-2\leq 0$

$\Leftrightarrow 2x(x-5)+(x-2)\leq 0$

Điều này vô lý do $2x(x-5)\geq 0; x-2\geq 3>0$ với mọi $x\geq 5$

Vậy pt vô nghiệm nên không có đáp án nào đúng.

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

ĐKXĐ: \(x\ne\pm2\)

\(\dfrac{x+1}{x-2}=\dfrac{2}{x^2-4}\)

\(\Rightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}=\dfrac{2}{x^2-4}\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)=2\)

\(\Leftrightarrow x^2+3x+2=2\)

\(\Leftrightarrow x^2+3x=0\)

\(\Leftrightarrow x\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\) (thỏa mãn)

đkxđ: \(x ≠2; x ≠-2\)

\(\dfrac{x+1}{x-2}=\dfrac{2}{x^2-4}\)

\(⇔\dfrac{(x+1)(x+2)}{x^2-4}=\dfrac{2}{x^2-4}\)

\(⇔(x+1)(x+2)=2\)

\(⇔x^2+3x=0\)

\(⇔x(x+3)=0\)

\(⇔\left[\begin{array}{} x=0\\ x+3=0 \end{array} \right.\)

\(⇔\left[\begin{array}{} x=0\\ x=-3 \end{array} \right.\)