\(A=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{2x-5}{x^2+5x}+\frac{x+3}{5-x}\)
Tìm x thuộc Z để A thuộc Z
GIẢI HỘ MIK vS!!
M=\(\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10}{1-6x+9x^2}\)
a, Tìm ĐKXĐ của M
b, Rút gọn M
Tính gtri của M vs x =\(\frac{1}{3}\)
CM biểu thức M k phụ thuộc vào x : P=\(\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{2x-5}{x^2+5x}+\frac{x}{5-x}\)
Giup mik vs . Mik đg cần gấp. Thanks
\(a.ĐKXĐ:\hept{\begin{cases}1-3x\ne0\\3x+1\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\...\\x\ge0\end{cases}}}\)
\(b,M=\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10}{1-6x+9x^2}\)
\(=\left(\frac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\frac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\left(\frac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\frac{5x+3x^2}{1+3x}.\frac{1-3x}{2\left(3x^2+5\right)}\)
==>Sai đề không mem
\(P=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{2x-5}{x^2+5x}+\frac{x}{5-x}\)
\(=\left(\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right):\frac{2x-5}{x\left(x+5\right)}+\frac{x}{5-x}\)
\(=\left(\frac{x^2}{x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{x\left(x+5\right)\left(x-5\right)}\right).\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)
\(=\frac{\left(x+x-5\right)\left(x-x+5\right)}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)
\(=\frac{5\left(2x-5\right)}{\left(x-5\right)}.\frac{1}{2x-5}+\frac{x}{5-x}\)
\(=\frac{5}{x-5}-\frac{x}{x-5}\)
\(=\frac{5-x}{x-5}\)
\(=\frac{-\left(x-5\right)}{x-5}\)
\(=-1\)
=> biểu thức P k phụ thuộc vào x
A=\(\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)
a, Rút gọn
b,Tìm x để A=2014
c,Tìm x ∈ z để A ∈ z
A=\(\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\)
a)rút gọn A
b)tìm x để A=-4
c)tính giá trị của A khi x2+4x+5
d)tìm x thuộc Z để A nhận giá trị nguyên
a)\(A=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\left(ĐK:x\ne0;-5\right)\)
\(\Leftrightarrow A=\frac{x^2}{5\left(x+5\right)}+\frac{2\left(x-5\right)}{x}+\frac{5\left(x+10\right)}{x\left(x+5\right)}\)
\(\Leftrightarrow A=\frac{x^3+10\left(x^2-25\right)+25x+250}{5x\left(x+5\right)}\)
\(\Leftrightarrow A=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)
\(\Leftrightarrow A=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}\)
\(\Leftrightarrow A=\frac{x+5}{5}\)
b)Để A=-4 \(\Leftrightarrow\frac{x+5}{5}=-4\)
\(\Leftrightarrow x+5=-20\)
\(\Leftrightarrow x=-25\)
a).....
\(=\frac{x^2}{5\left(x+5\right)}+\frac{2x-10}{x}+\frac{50+5x}{x\left(x+5\right)}\) MTC= 5x (x+5) ĐK\(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
\(=\frac{x^2.x}{5x\left(x+5\right)}+\frac{5.\left(2x-10\right).\left(x+5\right)}{5x\left(x+5\right)}+\frac{5.\left(50+5x\right)}{5x\left(x+5\right)}\)
\(=\frac{x^3+\left(10x-50\right).\left(x+5\right)+250+25x}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2+50x-50x-250+250+25x}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)
\(=\frac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)
\(=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)
b) A=-4
=>\(\frac{x+5}{5}=-4\)
=> x = -25
c)
d) Để A đạt gt nguyên thì 5\(⋮\)x+5
=> \(\left(x+5\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
*x+5=1 => x=-4 \(\in Z\)
*x+5=-1 => x=-6\(\in Z\)
*x+5=5 => x=0\(\in Z\)
*x+5=-5 => x=-10\(\in Z\)
Vậy...........
\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\left(\frac{x^2-2x}{x^3-x^2+x}\right)\))
a) Rút gọn
b) Tính giá trị A biết\(|x-\frac{3}{4}|=\frac{5}{4}\)
c) Tìm x thuộc Z để A thuộc Z
\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right)\div\left(\frac{x^2-2x}{x^3-x^2+x}\right)\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right)\div\left(\frac{x\left(x-2\right)}{x\left(x^2-x+1\right)}\right)\)
\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right)\div\frac{x-2}{x^2-x+1}\)
\(=\left(\frac{x+1+x+1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}\right)\times\frac{x^2-x+1}{x-2}\)
\(=\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}\times\frac{x^2-x+1}{x-2}\)
\(=\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{-2x}{x+1}\)
b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)
<=> \(\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=-\frac{1}{2}\left(nhan\right)\end{cases}}\)
Với x = -1/2 => \(A=\frac{-2\cdot\left(-\frac{1}{2}\right)}{-\frac{1}{2}+1}=2\)
c) Để A ∈ Z thì \(\frac{-2x}{x+1}\)∈ Z
=> -2x ⋮ x + 1
=> -2x - 2 + 2 ⋮ x + 1
=> -2( x + 1 ) + 2 ⋮ x + 1
Vì -2( x + 1 ) ⋮ ( x + 1 )
=> 2 ⋮ x + 1
=> x + 1 ∈ Ư(2) = { ±1 ; ±2 }
x+1 | 1 | -1 | 2 | -2 |
x | 0 | -2 | 1 | -3 |
Các giá trị trên đều tm \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
Vậy x ∈ { -3 ; -2 ; 0 ; 1 }
\(A=\left(\frac{x}{y^2-25}-\frac{x-5}{x^2+5x}\right):\frac{2x-5}{x^2+5x}+\frac{x}{5-x}\)
chứng minh rằng giá trị của A không phụ thuộc vào giá trị của x
\(A=\left(\frac{x^2-2x}{2x^2+8}-\frac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\)
Tìm x thuộc Z để A thuộc Z
cho \(P=\left(\frac{1}{x-2}-\frac{2x}{4-x^2}+\frac{1}{2+x}\right)\left(\frac{2}{x}-1\right)\)
a)rút gọn P
b)tìm x để p =1/2
c) tìm x thuộc z để P nhận gtri nguyên
giải hộ e vs ạ
\(ĐKXĐ:x\ne2;x\ne-2;x\ne0\)
\(a,P=\left(\frac{-1}{2-x}-\frac{2x}{4-x^2}+\frac{1}{2+x}\right)\left(\frac{2}{x}-1\right)\)
\(P=\left(\frac{-2-x+2-x-2x}{\left(2-x\right)\left(2+x\right)}\right)\left(\frac{2-x}{x}\right)\)
\(P=\frac{-4x}{\left(2-x\right)\left(2+x\right)}\frac{2-x}{x}\)
\(P=\frac{-4}{2+x}\)
\(b,P=\frac{-4}{2+x}=\frac{1}{2}\)
\(2+x=-8\)
\(x=-10\)
\(c,P=-\frac{4}{2+x}\)
\(< =>-4⋮x+2\)
lập bảng ra thì bạn ra đc \(x=\left\{2;-1;-3;-6\right\}\)
a)\(P=\left(\frac{1}{x-2}-\frac{2x}{4-x^2}+\frac{1}{2+x}\right)\left(\frac{2}{x}-1\right)\)
\(P=\left(\frac{1}{x-2}+\frac{2x}{\left(x+2\right)\left(x-2\right)}+\frac{1}{2+x}\right).\frac{2-x}{x}\)
\(P=\frac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}.\frac{2-x}{x}\)
\(P=\frac{4x}{\left(x-2\right)\left(x+2\right)}.\frac{2-x}{x}\)
\(P=\frac{-4}{x+2}\)
b) Để P=1/2
\(\Rightarrow-\frac{4}{x+2}=\frac{1}{2}\)
\(\Leftrightarrow-8=x+2\)
\(\Leftrightarrow x=-10\)
c) Để P nhận GT nguyên
\(\Rightarrow\left(x+2\right)\inƯ_{\left(-4\right)}\)
\(\Rightarrow\left(x+2\right)\in\left\{-1;1;-2;2;-4;4\right\}\)
\(\Rightarrow x=\left\{-3;-1;-4;0;-6;2\right\}\)
#H
Cho A=\(\left(\frac{3x}{x-2}-\frac{2x^2-5}{x^2-4}-\frac{x-1}{x+2}\right):\frac{3}{x+2}\)
a. Rút gọn A
b. Tính A biết \(x^2-2x=0\)
c. Tìm x thuộc Z để A thuộc Z
Cho A=\(\left(\frac{3x}{x-2}-\frac{2x^2-5}{x^2-4}-\frac{x-1}{x+2}\right):\frac{3}{x+2}\)
a. Rút gọn A
b. Tính A biết \(x^2-2x=0\)
c. Tìm x thuộc Z để A thuộc Z