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1b)
Đặt \(\overline{abcd}=k^2\left(k\in N;32\le k\le99\right)\)
Note : nếu k nằm ngoài khoảng giá trị ở trên thì k2 sẽ có ít hơn hoặc nhiều hơn 4 chữ số
Theo bài cho :
\(\overline{ab}-\overline{cd}=1\Rightarrow\overline{ab}=\overline{cd}+1\Rightarrow\overline{abcd}=k^2\Leftrightarrow100\cdot\overline{ab}+\overline{cd}=k^2\)
\(\Leftrightarrow100\cdot\overline{cd}+100+\overline{cd}=k^2\Leftrightarrow101\cdot\overline{cd}=k^2-100\Leftrightarrow101\overline{cd}=\left(k-10\right)\left(k+10\right)\)
\(\Rightarrow\orbr{\begin{cases}k-10⋮101\\k+10⋮101\end{cases}}\)
Mà \(\text{ }(k-10;101)=1\Rightarrow k+10⋮101\)
Lại có : \(32\le k\le99\Rightarrow42\le k+10\le109\)
\(\Rightarrow k+10=101\Rightarrow k=91\Rightarrow\overline{abcd}=91^2=8182\left(tm\right)\)
ta có :
\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)
\(\Leftrightarrow\left|x-1\right|\left|x+1\right|-\left|x-1\right|-\left|x+1\right|+1=0\)
\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)
\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)
aGiải phương trình |x-1|+|x-2|=|2x-3|
b)Giải phương trình 1/(x−2 )+ 2/(x−3) − 3/(x−5) = 1/(x^2 −5x+6)
d: ĐKXĐ: \(x\notin\left\{2;-3\right\}\)
\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)
=>\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{-5}{\left(x+3\right)\left(x-2\right)}\)
=>\(x+3-6\left(x-2\right)=-5\)
=>x+3-6x+12=-5
=>-5x+15=-5
=>-5x=-20
=>x=4(nhận)
e: ĐKXĐ: x<>-2
\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)
=>\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{5}{x^2-2x+4}\)
=>\(2\left(x^2-2x+4\right)-2x^2-16=5\left(x+2\right)\)
=>\(2x^2-4x+8-2x^2-16=5x+10\)
=>5x+10=-4x-8
=>9x=-18
=>x=-2(loại)
f: ĐKXĐ: \(x\in\left\{1;-1\right\}\)
\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)
\(\Leftrightarrow\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
=>\(\dfrac{\left(x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
=>\(\left(x^3+1\right)\left(x^2-1\right)-\left(x^3-1\right)\left(x^2-1\right)=2\left(x^2+4x+4\right)\)
=>\(\left(x^2-1\right)\cdot\left(x^3+1-x^3+1\right)=2\left(x^2+4x+4\right)\)
=>\(2x^2+8x+8=\left(x^2-1\right)\cdot2=2x^2-2\)
=>8x=-10
=>x=-5/4(nhận)
1: Ta có: \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\)
Suy ra: \(x^2+4x+4+2x-4=x^2\)
\(\Leftrightarrow6x=0\)
hay \(x=0\left(nhận\right)\)
2: Ta có: \(\dfrac{1}{x-6}-\dfrac{2}{x+6}=\dfrac{3x+6}{x^2-36}\)
Suy ra: \(x+6-2x+12=3x+6\)
\(\Leftrightarrow-x-3x=6-18=-12\)
hay \(x=3\left(nhận\right)\)
Lời giải:
1. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{(x+2)^2+2(x-2)}{(x-2)(x+2)}=\frac{x^2}{x^2-4}\)
\(\Leftrightarrow \frac{x^2+6x}{x^2-4}=\frac{x^2}{x^2-4}\)
\(\Rightarrow x^2+6x=x^2\Leftrightarrow x=0\) (tm)
2. ĐKXĐ: $x\neq \pm 6$
PT \(\Leftrightarrow \frac{6+x-2(x-6)}{(x-6)(6+x)}=\frac{3x+6}{x^2-36}\)
\(\Leftrightarrow \frac{18-x}{x^2-36}=\frac{3x+6}{x^2-36}\)
\(\Rightarrow 18-x=3x+6\Leftrightarrow 12=4x\Leftrightarrow x=3\) (tm)
=>x^2 - 4xy+4y^2+y^2 = 13^2
=>(x-2y)^2 + y^2 - 13^2 =0
=> x - 2y + y - 13 = 0
=> x = y + 13