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\(\left(x-1\right)-\left(x-2\right)\left(x+2\right)\)
\(=\left(x-1\right)-\left(x^2-2^2\right)\)
\(=\left(x-1\right)-x^2+2^2\)
\(=x-1-x^2+2^2\)
\(=x-x^2+\left(2-1\right)\left(2+1\right)\)
\(=x-x^2+3\)
a/ (x-1)2-(x-2)(x+2)
=(x-1)-(x2-22)
=(x-1)-x2-22
=x-x2 +(2-1)(2+1)
=x-x2+3
\(ĐK:x\ne\pm3\)
\(P=\left[\frac{\left(2x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-10x}{\left(x-3\right)\left(x+3\right)}\right]\cdot\frac{x-3}{x+2}\)
\(=\frac{2x^2-7x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x-3}{x+2}\)
\(=\frac{3x^2+6x}{x+3}\cdot\frac{1}{x+2}=\frac{3x\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}=\frac{3x}{x+3}\)
\(A=\frac{\left[x\left(x^2-x+1\right)\right]-\left[\left(x+1\right)\left(3-3x\right)\right]+\left[x+4\right]}{x^3+1}\)
\(A=\frac{\left(x^3-x^2+x\right)+3\left(x^2-1\right)+\left(x+4\right)}{x^3+1}=\frac{x^3+2x^2+2x+1}{x^3+1}\)
\(A=\frac{\left(x^3+1\right)+2x\left(x+1\right)}{x^3+1}=1+\frac{2x}{x^2-x+1}\)
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x\left(x^2-x+1\right)-\left(3+3x\right)\left(x+1\right)+\left(x+4\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2+x-9x-3-3x^2+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2-3x^2+x-9x+x+3+4}{x^3+1}\)
\(A=\frac{x^3+2x^2-4x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(B=\left(\frac{2x}{x-3}-\frac{x-1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\left(ĐK:x\ne\pm3\right)\)
\(=\frac{2x\left(x+3\right)-\left(x-1\right)\left(x-3\right)-x^2-1}{x^2-9}:\frac{x+3-x+1}{x+3}\)
\(=\frac{2x^2+6x-x^2+3x+x-3-x^2-1}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{4}\)
\(=\frac{10x-4}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{4}=\frac{10x-4}{4\left(x-3\right)}\)
\(B=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)
\(=\left[\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}\right]:\left(\frac{x+3-x+1}{x+3}\right)\)
\(=\left(\frac{2x^2+6x-x^2+3x-x+3-x^2-1}{\left(x+3\right)\left(x-3\right)}\right):\frac{4}{x+3}\)
\(=\frac{8x-1}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{4}\)\(=\frac{8x-1}{4\left(x-3\right)}\)
\(B=\left(\frac{21}{x^2-9}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+2}{x+3}\)
\(B=\frac{2x^2-5x+12}{x^2-9}\cdot\frac{x+3}{x+2}\)
\(B=\frac{2x^2-5x-12}{\left(x-3\right)\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{x^2-x-6}\)
Thik thì tách tiếp nha
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
1. P = \(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\) ĐKXĐ: \(x\ne-3\), \(x\ne2\)
= \(\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)
= \(\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{x-2}\)
= \(\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
= \(\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
= \(\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
= \(\frac{x-4}{x-2}\)
2. P=\(\frac{-3}{4}\)
<=> \(\frac{x-4}{x-2}=\frac{-3}{4}\)
<=> 4 ( x - 4 ) = -3 ( x - 2 )
<=> 4x - 16 = -3x + 6
<=> 7x = 2
<=> x = \(\frac{22}{7}\)
3. \(x^2-9=0\)
<=> ( x -3 ) ( x + 3 ) = 0
<=> \(\orbr{\begin{cases}x=3\left(tm\right)\\x=-3\left(ktm\right)\end{cases}}\)
-> P = \(\frac{3-4}{3-2}\) = -1
ĐKXĐ : \(x\ne\pm3\)
a) \(A=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)
\(A=\left(\frac{-2x\left(3+x\right)}{\left(3-x\right)\left(3+x\right)}-\frac{\left(x+1\right)\left(3-x\right)}{\left(x+3\right)\left(3-x\right)}+\frac{x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3}{x+3}-\frac{x-1}{x+3}\right)\)
\(A=\left(\frac{-2x^2-6x+x^2-2x-3+x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3-x+1}{x+3}\right)\)
\(A=\left(\frac{-8x-2}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{4}{x+3}\right)\)
\(A=\frac{-2\left(4x+1\right)\left(x+3\right)}{\left(3-x\right)\left(3+x\right)4}\)
\(A=\frac{-\left(4x+1\right)}{2\left(3-x\right)}\)
\(A=\frac{4x+1}{2\left(x-3\right)}\)
b) \(\left|x-5\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}}\)
Mà ĐKXĐ x khác 3 => ta xét x = 7
\(A=\frac{4\cdot7+1}{2\cdot\left(7-3\right)}=\frac{29}{8}\)
c) Để A nguyên thì 4x + 1 ⋮ 2x - 3
<=> 4x - 6 + 7 ⋮ 2x - 3
<=> 2 ( 2x - 3 ) + 7 ⋮ 2x - 3
Mà 2 ( 2x - 3 ) ⋮ ( 2x - 3 ) => 7 ⋮ 2x - 3
=> 2x - 3 thuộc Ư(7) = { 1; -1; 7; -7 }
=> x thuộc { 2; 1; 5; -2 }
Vậy .....
a) ĐKXĐ: \(x\ne\pm3\)
\(A=\frac{2x\left(x+3\right)-\left(x+1\right)\left(x-3\right)-\left(x^2+1\right)}{x^2-9} : \frac{x+3-\left(x-1\right)}{x+3}\)
\(A=\frac{2x^2-6x-x^2+2x+3-x^2-1}{x^2-9} : \frac{4}{x+3}\)
\(A=\frac{-4x+2}{x^2+9} : \frac{4}{x+3}\)
\(A=\frac{2\left(1-2x\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{4}=\frac{1-2x}{2x-6}\)
b)
Có 2 trường hợp:
T.Hợp 1:
\(x-5=2\Leftrightarrow x=7\)(thỏa mã ĐKXĐ)
thay vào A ta được: A=\(-\frac{13}{8}\)
T.Hợp 2:
\(x-5=-2\Leftrightarrow x=3\)(Không thỏa mãn ĐKXĐ)
Vậy không tồn tại giá trị của A tại x=3
Vậy với x=7 thì A=-13/8
c)
\(\frac{1-2x}{2x-6}=\frac{1-\left(2x-6\right)-6}{2x-6}=-1-\frac{5}{2x-6}\)
Do -1 nguyên, để A nguyên thì \(-\frac{5}{2x-6}\inℤ\)
Để \(-\frac{5}{2x-6}\inℤ\)thì \(2x-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Do 2x-6 chẵn, để x nguyên thì 2x-6 là 1 số chẵn .
Vậy không có giá trị nguyên nào của x để A nguyên
\(\left(x^2+\frac{1}{x}+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3\)
\(=\left[x^3-\left(\frac{1}{3}\right)^3\right]-\left(x-\frac{1}{3}\right)^3\)
\(=\left(x-\frac{1}{3}\right)^3-\left(x-\frac{1}{3}\right)^3\)
\(=\left(x-\frac{1}{3}\right)\left[x^2+\frac{1}{x}+\frac{1}{9}-\left(x-\frac{1}{3}\right)^2\right]\)
\(=\left(x-\frac{1}{3}\right)\left(\frac{1}{x}+\frac{2x}{3}\right)\)
\(=\frac{3x-1}{3}\times\frac{3+2x^2}{3x}\)
\(=\frac{9x+6x^2-3-2x^2}{9x}\)
\(=\frac{4x^2+9x-3}{9x}\)