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Đề bài là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}\) hay là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2}-\left(x+2\right)^2?\)
\(\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}\)
viết lại biểu thức
1. ĐKXĐ: \(x\ne\pm1\)
2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)
\(=\dfrac{\left(x+1\right)^2-\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{x^2+2x+1-x^2+4x-3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x-3}{x-1}\)
3. Tại x = 5, A có giá trị là:
\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)
4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)
Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)
Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)
1.a)\(\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3}{\left(x+2\right)\left(x-2\right)}-\frac{x}{x-2}-\frac{2}{x+2}\)
Để biểu thức được xác định thì:\(\left(x+2\right)\left(x-2\right)\ne0\)\(\Rightarrow x\ne\pm2\)
\(\left(x+2\right)\ne0\Rightarrow x\ne-2\)
\(\left(x-2\right)\ne0\Rightarrow x\ne2\)
Vậy để biểu thức xác định thì : \(x\ne\pm2\)
b) để C=0 thì ....
1, c , bn Nguyễn Hữu Triết chưa lm xong
ta có : \(/x-5/=2\)
\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}\)
thay x = 7 vào biểu thứcC
\(\Rightarrow C=\frac{4.7^2\left(2-7\right)}{\left(7-3\right)\left(2+7\right)}=\frac{-988}{36}=\frac{-247}{9}\)KL :>...
thay x = 3 vào C
\(\Rightarrow C=\frac{4.3^2\left(2-3\right)}{\left(3-3\right)\left(3+7\right)}\)
=> ko tìm đc giá trị C tại x = 3
a. ĐKXĐ: \(x\ne\pm1\)
b. \(A=\left(x^2-1\right)\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left[\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right]\)
\(=\left(x-1\right)\left(x+1\right)\left[\dfrac{x+1-x+1-\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right]\)
\(=\left(x-1\right)\left(x+1\right)\left[\dfrac{-x^2+3}{\left(x-1\right)\left(x+1\right)}\right]\)
\(=\dfrac{\left(x-1\right)\left(x+1\right)\left(-x^2+3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=-x^2+3\)
c. Thay x = 3 vào A ta được:
\(-\left(3\right)^2+3=-6\)
Vậy: Giá trị của A tại x = 3 là -6
a) ĐKXĐ: \(x\ne1;x\ne-1.\)
b) \(A=\left(x^2-1\right).\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}-1\right).\)
\(=\left(x^2-1\right).\dfrac{x+1-x+1-x^2+1}{x^2-1}=-x^2+3.\)
c) Thay x = 3 (TMĐK) vào A: \(-3^2+3=-6.\)
a) đk: x khác 0;1
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
= \(\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\right]\)
= \(\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
= \(\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b) Để \(\left|2x-5\right|=3\)
<=> \(\left[{}\begin{matrix}2x-5=3< =>2x=8< =>x=4\left(c\right)\\2x-5=-3< =>2x=2< =>x=1\left(l\right)\end{matrix}\right.\)
Thay x = 4 vào A, ta có:
\(A=\dfrac{4^2}{4-1}=\dfrac{16}{3}\)
c) Để A = 4
<=> \(\dfrac{x^2}{x-1}=4\)
<=> \(\dfrac{x^2}{x-1}-4=0< =>\dfrac{x^2-4x+4}{x-1}=0\)
<=> \(\left(x-2\right)^2=0\)
<=> x = 2 (T/m)
d) Để A < 2
<=> \(\dfrac{x^2}{x-1}< 2< =>\dfrac{x^2}{x-1}-2< 0< =>\dfrac{x^2-2x+2}{x-1}< 0\)
<=> \(\dfrac{\left(x-1\right)^2+1}{x-1}< 0\)
Mà \(\left(x-1\right)^2+1>0\)
<=> x - 1 < 0 <=> x < 1
KHĐK: x < 1 ( x khác 0)
e) Để A thuộc Z
<=> \(\dfrac{x^2}{x-1}\in Z\)
<=> \(x^2⋮x-1\)
<=> \(x^2-x\left(x-1\right)-\left(x-1\right)⋮x-1\)
<=> \(1⋮x-1\)
Ta có bảng:
| x-1 | 1 | -1 |
| x | 2 | 0 |
| T/m | T/m |
KL: Để A thuộc Z <=> \(x\in\left\{2;0\right\}\)
f) Để A thuộc N <=> \(x\in\left\{2;0\right\}\)
a)\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
A xác định
\(\Leftrightarrow\hept{\begin{cases}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-3\\\left(x+3\right)\left(x-2\right)\ne0\\x\ne2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)
Vậy A xác định \(\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)
b) \(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x^2-2x\right)+\left(3x-6\right)}+\frac{1}{2-x}\)
\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{x.\left(x-2\right)+3.\left(x-2\right)}+\frac{1}{2-x}\)
\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}+\frac{1}{2-x}\)
\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)
\(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(A=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{\left(x^2+3x\right)-\left(4x+12\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x.\left(x+3\right)-4.\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x-4}{x-2}\left(x+3\ne0\right)\)
c) \(A=-\frac{3}{4}\)
\(\Leftrightarrow\frac{x-4}{x-2}=-\frac{3}{4}\)
\(\Leftrightarrow4.\left(x-4\right)=-3.\left(x-2\right)\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow7x=22\)
\(\Leftrightarrow x=\frac{22}{7}\)
Vậy \(x=\frac{22}{7}\)
Tham khảo nhé~
a) ĐKXĐ: x∉{2;-3}
Ta có: \(A=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-4-5}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-9}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+3\right)\left(x-3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-3}{x-2}\)
b) Thay x=-2 vào biểu thức \(A=\frac{x-3}{x-2}\), ta được:
\(\frac{-2-3}{-2-2}=\frac{-5}{-4}=\frac{5}{4}\)
Vậy: \(\frac{5}{4}\) là giá trị của biểu thức \(A=\frac{x-3}{x-2}\) tại x=-2
c) Đặt A=5
⇒\(\frac{x-3}{x-2}=5\)
\(\Leftrightarrow x-3=5\left(x-2\right)\)
\(\Leftrightarrow x-3=5x-10\)
\(\Leftrightarrow x-3-5x+10=0\)
\(\Leftrightarrow-4x+7=0\)
\(\Leftrightarrow-4x=-7\)
hay \(x=\frac{7}{4}\)(tm)
Vậy: Khi \(x=\frac{7}{4}\) thì A=5
Đặt A=0
⇒\(\frac{x-3}{x-2}=0\)
⇔x-3=0
hay x=3(tm)
Vậy: Khi x=3 thì A=0
d) Để biểu thức A có giá trị nguyên thì x-3⋮x-2
⇔x-2-1⋮x-2
mà x-2⋮x-2
nên -1⋮x-2
⇔x-2∈Ư(-1)
⇔x-2∈{1;-1}
hay x∈{3;1}(tm)
Vậy: Khi x∈{1;3} thì A có giá trị nguyên