Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, sửa đề : \(C=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{1}{2-x}\)ĐK : \(x\ne-3;2\)
\(=\frac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-12-x}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b, Ta có : \(x^2-x=2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\Leftrightarrow x=-1;x=2\)
Kết hợp với giả thiết vậy x = -1
Thay x = -1 vào biểu thức C ta được : \(\frac{-1-4}{-1-2}=-\frac{5}{-3}=\frac{5}{3}\)
c, Ta có : \(C=\frac{1}{2}\Rightarrow\frac{x-4}{x-2}=\frac{1}{2}\Rightarrow2x-8=x-2\Leftrightarrow x=6\)( tm )
d, \(C>1\Rightarrow\frac{x-4}{x-2}>1\Rightarrow\frac{x-4}{x-2}-1>0\Leftrightarrow\frac{x-4-x+2}{x-2}>0\Leftrightarrow\frac{-2}{x-2}>0\)
\(\Rightarrow x-2< 0\Leftrightarrow x< 2\)vì -2 < 0
e, tự làm nhéee
f, \(C< 0\Rightarrow\frac{x+4}{x+2}< 0\)
mà x + 4 > x + 2
\(\hept{\begin{cases}x+4>0\\x+2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-4\\x< -2\end{cases}\Leftrightarrow-4< x< -2}}\)
Vì \(x\inℤ\Rightarrow x=-3\)( ktmđk )
Vậy ko có x nguyên để C < 0
g, Ta có : \(\frac{x+4}{x+2}=\frac{x+2+2}{x+2}=1+\frac{2}{x+2}\)
Để C nguyên khi \(x+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x + 2 | 1 | -1 | 2 | -2 |
| x | -1 | -3 | 0 | -4 |
h, Ta có : \(D=C\left(x^2-4\right)=\frac{x+4}{x+2}.\frac{\left(x-2\right)\left(x+2\right)}{1}=x^2+2x-8\)
\(=\left(x+1\right)^2-9\ge-9\)
Dấu ''='' xảy ra khi x = -1
Vậy GTNN D là -9 khi x = -1
a) Đk: x > 0 và x khác +-1
Ta có: A = \(\left(\frac{x+1}{x}-\frac{1}{1-x}-\frac{x^2-2}{x^2-x}\right):\frac{x^2+x}{x^2-2x+1}\)
A = \(\left[\frac{\left(x-1\right)\left(x+1\right)+x-x^2+2}{x\left(x-1\right)}\right]:\frac{x\left(x+1\right)}{\left(x-1\right)^2}\)
A = \(\frac{x^2-1+x-x^2+2}{x\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{x\left(x+1\right)}\)
A = \(\frac{x+1}{x}\cdot\frac{x-1}{x\left(x+1\right)}=\frac{x-1}{x^2}\)
b) Ta có: A = \(\frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}=-\left(\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\frac{1}{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> 1/x - 1/2 = 0 <=> x = 2 (tm)
Vậy MaxA = 1/4 <=> x = 2
a) \(A=\left(\frac{2+x}{2-x}-\frac{2-x}{2+x}-\frac{4x^2}{x^2-4}\right):\frac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)(ĐKXĐ: \(\hept{\begin{cases}x\ne\pm2\\x\ne3\end{cases}}\))\(=\left[\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{4-x^2}\right]:\frac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\)\(=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}.\frac{2-x}{x-3}=\frac{4x}{x-3}\)
b) l\(x-5\)l\(=2\Leftrightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\left(n\right)\\x=3\left(l\right)\end{cases}\Rightarrow A=\frac{4.7}{7-3}=\frac{28}{4}=7}\)
c)
* Để A có giá trị là một số nguyên thì \(A=\frac{4x}{x-3}=\frac{4x-12+12}{x-3}=4+\frac{12}{x-3}\)là một số nguyên hay \(\frac{12}{x-3}\)là một số nguyên \(\Rightarrow x-3\inƯ\left(12\right)\Rightarrow S=\left(-9;-3;-1;0;1;4;5;6;7;9;15\right)\)(1)
* Để \(A=4+\frac{12}{x-3}< 4\Leftrightarrow\frac{12}{x-3}< 0\) thì \(x-3< 0\Leftrightarrow x< 3\)(2)
(1)(2) \(\Rightarrow S=\left(-9;-3;-1;0;1\right)\)
Mình nghĩ bạn viết hơi sai đề bài.
\(x^2+xz-y^2-yz=\left(x^2-y^2\right)+xz-yz=\left(x-y\right)\left(x+y\right)+z\left(x-y\right)=\left(x-y\right)\left(x+y+z\right)\)
Tương tự: \(y^2+xy-z^2-xz=\left(y-z\right)\left(x+y+z\right)\)
\(z^2+yz-x^2-xy=\left(x+y+z\right)\left(z-x\right)\)
Khi đó:
\(P=\frac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}+\frac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}+\frac{1}{\left(x-y\right)\left(x+y+z\right)\left(z-x\right)}\)
\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}=0\)
a) Phân thức M xác định khi và chỉ khi :
+) \(2x-2\ne0\Leftrightarrow x\ne1\)
+) \(2x+2\ne0\Leftrightarrow x\ne-1\)
+) \(1-\frac{x-3}{x+1}\ne0\)
\(\Leftrightarrow x-3\ne x+1\)
\(\Leftrightarrow0x\ne4\left(\text{luôn đúng}\right)\)
Vậy \(x\ne\left\{1;-1\right\}\)
b) \(M=\left(\frac{x-2}{2x-2}-\frac{x+3}{2x+2}+\frac{3}{2x-2}\right):\left(1-\frac{x-3}{x+1}\right)\)
\(M=\left(\frac{\left(x-2\right)\left(2x+2\right)}{\left(2x-2\right)\left(2x+2\right)}-\frac{\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}+\frac{3\left(2x+2\right)}{\left(2x-2\right)\left(2x+2\right)}\right):\left(\frac{x+1-x+3}{x+1}\right)\)
\(M=\left(\frac{2x^2-2x-4-2x^2-4x+6+6x+6}{\left(2x-2\right)\left(2x+2\right)}\right):\left(\frac{4}{x+1}\right)\)
\(M=\frac{8}{2\left(x-1\right)2\left(x+1\right)}\cdot\frac{x+1}{4}\)
\(M=\frac{8\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)\cdot4}\)
\(M=\frac{8\left(x+1\right)}{8\left(x+1\right)\left(x-1\right)}\)
\(M=\frac{1}{x-1}\)
\(M=\left(\frac{x-2}{2x-2}-\frac{x+3}{2x+2}+\frac{3}{2x-2}\right):\left(1-\frac{x-3}{x+1}\right)\)
\(=\left(\frac{x+1}{2x-2}-\frac{x+3}{2x+2}\right):\left(\frac{4}{x+1}\right)=\left[\frac{\left(x+1\right)\left(2x+2\right)-\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]:\left(\frac{4}{x+1}\right)\)
\(=\left[\frac{2x^2+4x+2-2x^2+2x+6-6x+6}{4x^2-4}\right]:\left(\frac{4}{x+1}\right)\)
\(=\left[\frac{6x+8-6x+6}{4x^2-4}\right]:\left(\frac{4}{x+1}\right)\)
\(=\frac{14}{4x^2-4}:\left(\frac{4}{x+1}\right)=\frac{14x+14}{16x^2-16}=\frac{7x+7}{8x^2-8}\)
d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
a) \(ĐKXĐ:\hept{\begin{cases}x^3+1\ne0\\x^3-2x^2\ne0\\x+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne2\end{cases}}\)(chỗ chữ và là do OLM thiếu ngoặc 4 cái nên mk để thế nha! trình bày thì kẻ thêm 1 ngoặc nưax)
\(Q=1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left[\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right]:\frac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(=1+\frac{\left(x+1\right)+\left(x+1\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\frac{4x-2x^2}{x+1}.\frac{1}{x\left(x-2\right)}\)
\(=1-\frac{2x\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}=1-\frac{2}{x+1}=\frac{x-1}{x+1}\)
b, Với \(x\ne0;x\ne-1;x\ne2\)Ta có:
\(|x-\frac{3}{4}|=\frac{5}{4}\)
*TH1:
\(x-\frac{3}{4}=\frac{5}{4}\Rightarrow x=2\)(ko thảo mãn)
*TH2:
\(x-\frac{3}{4}=-\frac{5}{4}\Rightarrow x=-\frac{1}{2}\)
\(\Rightarrow Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
c,
\(Q=\frac{x-1}{x+1}=1-\frac{2}{x+1}\)
Để Q nguyên thì x+1 phải thuộc ước của 2!! tự làm tiếp dễ rồi!!
a)ĐKXĐ: x\(\ne\)1;x\(\ne\)-1
B=\(\frac{1}{4x-4}\)
b)
B=\(\frac{1}{8016}\)
c)
x=\(\frac{4007}{4008}\)
\(B=\left(\frac{x-4}{x\left(x-2\right)}+\frac{2}{x-2}\right):\left(\frac{x+2}{x}-\frac{x}{x-2}\right)\)
\(< =>B=\left(\frac{x-4}{x\left(x-2\right)}+\frac{2x}{x\left(x-2\right)}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}+\frac{x.x}{x\left(x-2\right)}\right)\)
\(< =>B=\left(\frac{x-4+2x}{x\left(x-2\right)}\right):\left(\frac{x^2-4}{x\left(x-2\right)}+\frac{x^2}{x\left(x-2\right)}\right)\)
\(< =>B=\frac{3x-4}{x\left(x-2\right)}:\frac{x^2-4+x^2}{x\left(x-2\right)}\)
\(< =>B=\frac{3x-4}{x\left(x-2\right)}.\frac{x\left(x-2\right)}{2x^2-4}\)
\(< =>B=\frac{3x-4}{2x^2-4}\)
\(b,\)Với \(x=-2\)thì
\(B=\frac{3\left(-2\right)-4}{2\left(-2\right)^2-4}=\frac{-6-4}{8-4}=-\frac{10}{4}=-\frac{5}{2}\)
\(ĐKXĐ:x\ne2;x\ne0\)
a
\(B=\left[\frac{x-4}{x\left(x-2\right)}+\frac{2}{x-2}\right]:\left(\frac{x+2}{x}-\frac{x}{x-2}\right)\)
\(=\frac{x-4+2x}{x\left(x-2\right)}:\frac{\left(x+2\right)\left(x-2\right)-x^2}{x\left(x-2\right)}\)
\(=\frac{3x-4}{x^2-4-x^2}=-\frac{3x-4}{4}\)
b
\(B=-\frac{3x-4}{4}=-\frac{3\cdot\left(-2\right)-4}{4}=\frac{5}{2}\)
c
\(\left|B\right|-2x=5\Leftrightarrow\left|B\right|=5+2x\)
\(B=-\frac{3x-4}{4}\Leftrightarrow-\frac{3x-4}{4}\ge0\Leftrightarrow x\le\frac{4}{3}\)
\(B=\frac{3x-4}{4}\Leftrightarrow x>\frac{4}{3}\)
Xét các trường hợp của x thì ra nghiệm bạn nhé
d
\(\left(2-x\right)B=-\frac{\left(2-x\right)\left(3x-4\right)}{4}\)
Để ( 2 - x ).B đạt giá trị nhỏ nhất thì ( 2 - x ) ( 3x - 4 ) đạt giá trị lớn nhất
Casio sẽ giúp chúng ta phần này
e
Để B là số nguyên âm lớn nhất hay \(B=-1\Leftrightarrow-\frac{3x-4}{4}=-1\Leftrightarrow x=\frac{8}{3}\)
g
\(\left|B\right|+3< 2x-1\)
Làm hệt như câu c nhé :D