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Bài 1:
ĐKXĐ: \(x\ne\left\{-1;1\right\}\)
\(P=\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right).\frac{4\left(x^2-1\right)}{5}\)
\(P=\left(\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x+3\right)}{2\left(x-1\right)\left(x+1\right)}\right).\frac{4\left(x^2-1\right)}{5}\)
\(P=\left(\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x^2-1\right)}\right)\frac{4\left(x^2-1\right)}{5}\)
\(P=\frac{10.4.\left(x^2-1\right)}{2\left(x^2-1\right).5}=\frac{40}{10}=4\)
Bài 2:
ĐK: \(x\ne\left\{-2;2;\right\}\)
\(A=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)
\(A=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right).\frac{x+2}{6}\)
\(A=\left(\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)}{6}\)
\(A=\frac{-6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}=\frac{-1}{x-2}\)
b/ \(\left|x\right|=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}A=\frac{-1}{\frac{1}{2}-2}=\frac{2}{3}\\A=\frac{-1}{-\frac{1}{2}-2}=\frac{2}{5}\end{matrix}\right.\)
c/ \(A< 0\Rightarrow\frac{-1}{x-2}< 0\Rightarrow\frac{1}{x-2}>0\Rightarrow x-2>0\Rightarrow x>2\)
\(\)
Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
3b. Để A=\(\frac{4x^3-6x^2+8x}{2x-1}\) \(\in\)Z => 2x2-2x+3+\(\frac{3}{2x-1}\)\(\in\)Z =>\(\frac{3}{2x-1}\) \(\in\)Z
=> 2x-1 \(\in\)Ư(3)={\(\pm\)1,\(\pm\)3}
=> \(\left[{}\begin{matrix}2x-1=1\\2x-1=-1\\2x-1=3\\2x-1=-3\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=0\\x=2\\x=-1\end{matrix}\right.\)(tm)
a) Ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)
\(\Rightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Rightarrow ab+bc+ac=0\)
Ta lại có:
\(a+b+c=1\)
\(\Rightarrow\left(a+b+c\right)^2=1\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=1\)
=> Đpcm
a, ĐKXĐ : \(\left\{{}\begin{matrix}x^3-1\ne0\\x^2+x+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x-1\ne0\\\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ne0\end{matrix}\right.\)
=> \(x\ne1\)
- Ta có : \(A=\left(\frac{x-1}{x^2+x+1}-\frac{x^2-3x+1}{x^3-1}-\frac{1}{x-1}\right):\frac{x^2+1}{1-x}\)
=> \(A=\left(\frac{\left(x-1\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}-\frac{x^2-3x+1}{x^3-1}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x^2+1}{1-x}\)
=> \(A=\left(\frac{x^2-2x+1-x^2+3x-1-x^2-x-1}{\left(x^2+x+1\right)\left(x-1\right)}\right):\frac{x^2+1}{1-x}\)
=> \(A=\left(\frac{-x^2-1}{\left(x^2+x+1\right)\left(x-1\right)}\right)\left(\frac{1-x}{x^2+1}\right)\)
=> \(A=\frac{-\left(x^2+1\right)\left(1-x\right)}{\left(x^2+x+1\right)\left(x-1\right)\left(x^2+1\right)}\)
=> \(A=\frac{1}{x^2+x+1}\)
b, Ta có : \(\frac{1}{A}=x^2+x+1\)
- Để \(\frac{1}{A}\) là số chính phương khi :
\(x^2+x+1=y^2\)
=> \(x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=y^2\)
=> \(\left(x+\frac{1}{2}\right)^2-y^2=-\frac{3}{4}\)
=> \(\left(x+\frac{1}{2}-y\right)\left(x+\frac{1}{2}+y\right)=-\frac{3}{4}\)
Vậy không tồn tại giá trị nguyên của x để \(\frac{1}{A}\) là số chính phương .
b)áp dụng Bđt cô si
\(\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge2\sqrt{\frac{x^2}{y^2}\cdot\frac{y^2}{x^2}}=2\)
\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}\cdot\frac{y}{x}}=2\)\(\Rightarrow-3\left(\frac{x}{y}+\frac{y}{x}\right)\ge-6\)
\(\Rightarrow P\ge2+\left(-5\right)+5=1\)
Dấu = khi x=y
a)Áp dụng Bđt Cô si ta có:
\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}\cdot\frac{y}{x}}=2\)
Dấu = khi \(x=y\)
Bài 1:
\(\frac{\frac{x}{x-y}-\frac{y}{x+y}}{\frac{y}{x-y}+\frac{x}{x+y}}=\frac{\frac{x(x+y)-y(x-y)}{(x-y)(x+y)}}{\frac{y(x+y+x(x-y)}{(x-y)(x+y)}}=\frac{\frac{x^2+y^2}{(x-y)(x+y)}}{\frac{x^2+y^2}{(x-y)(x+y)}}=1\)
Bài 2:
a)
ĐKXĐ: \(\left\{\begin{matrix} x-5\neq 0\\ x^2-25\neq 0\\ x+5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-5\neq 0\\ (x-5)(x+5)\neq 0\\ x+5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+5\neq 0\\ x-5\neq 0\end{matrix}\right.\Leftrightarrow x\neq \pm 5\)
b)
\(A=\frac{x(x+5)}{(x+5)(x-5)}-\frac{10x}{(x-5)(x+5)}-\frac{5(x-5)}{(x-5)(x+5)}=\frac{x(x+5)-10x-5(x-5)}{(x-5)(x+5)}\)
\(=\frac{x^2-10x+25}{(x-5)(x+5)}=\frac{(x-5)^2}{(x-5)(x+5)}=\frac{x-5}{x+5}\)
c)
Khi $x=9$ thì $A=\frac{9-5}{9+5}=\frac{2}{7}$
c/\(Q=\frac{1}{x^2-x+1}\)
Ta có \(x^2-x+1=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Suy ra MAX Q=3/4 với x=1/2
\(\)Câu 1 ĐK : x khác -1
a ) \(A=\frac{3x+3}{x^3+x^2+x+1}=\frac{3\left(x+1\right)}{\left(x^2+1\right)\left(x+1\right)}=\frac{3}{x^2+1}\)
b) Thiếu đề , đề phải là x nguyên
=> \(3⋮x^2+1\Rightarrow x^2+1\in\left\{\pm1;\pm3\right\}\)
Mà x nguyên nên x \(\in\left\{-2;0\right\}\)
c) Ta có \(x^2+1\ge1\Rightarrow\frac{3}{x^2+1}>0\)
=> Phân thức đạt giá trị lớn nhất khi \(x^2+1\) nhỏ nhất
=> x = 0
=> GTLN của A = \(\frac{3}{1}=3\)
Câu 2
a ) \(\left|x-4\right|+\left|x-12\right|=8\) (*)
Vời \(x\ge12\)
Phương trình (*) tương đương
x -4 + x -12 = 8
=> 2x -16 =8
=> 2x = 24
=>x = 12 (1)
Với \(4\le x< 12\) có
(* ) tương đương
x -4 +12 - x = 8
=> 8 = 8
=> PT có nghiệm \(4\le x< 12\) (2)
Với \(x< 4\) , có (*) tương đương
4-x +12 - x = 0
=> 16 - 2x = 0
=> x = 8 (3)
Kết hợp (1); (2) ;(3) có x là nghiệm của phương trình với \(4\le x\le12\)