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thiếu đề : \(\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}.\)
Bài 2 :
a, Để \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4^2-4}{5}\)
\(\Rightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}}\Rightarrow\orbr{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)
b,\(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4x^2-4}{5}\)
\(B=\left[\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x+1\right)\left(x-1\right)}-\frac{x+3}{2\left(x+1\right)}\right].\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\left[\frac{x^2+2x+1}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\left[\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\frac{4}{2\left(x-1\right)\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\frac{8}{5}\)
=> giá trị của B ko phụ thuộc vào biến x
bài 1
=\(^{\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x+1\right)^2}\)
=\(\left(2x+1+2x-1\right)^2\)
=\(\left(4x\right)^2\)
=\(16x^2\)
Tại x=100 thay vào biểu thức trên ta có:
16*100^2=1600000
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
Bài 1:
a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)
\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)
Suy ra: \(12x-45-12x^2+45x=0\)
\(\Leftrightarrow-12x^2+57x-45=0\)
\(\Leftrightarrow-12x^2+12x+45x-45=0\)
\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)
\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)
mà \(-3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)
b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)
Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)
\(\Leftrightarrow-x^2+16x-39=0\)
\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)
\(\Leftrightarrow x^2-13x-3x+39=0\)
\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)
Vậy: Tập nghiệm S={3;13}
c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)
\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)
Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)
\(\Leftrightarrow-21x^2+26x+11=0\)
\(\Leftrightarrow-21x^2-7x+33x+11=0\)
\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
a) \(ĐKXĐ:x\ne1\)
b) \(\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right):\left(1-\frac{2x}{x^2+1}\right)\)
\(=\left(\frac{1}{x-1}-\frac{2x}{x\left(x^2+1\right)-\left(x^2+1\right)}\right):\frac{x^2+1-2x}{x^2+1}\)
\(=\left(\frac{1}{x-1}-\frac{2x}{\left(x^2+1\right)\left(x-1\right)}\right):\frac{\left(x-1\right)^2}{x^2+1}\)
\(=\frac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}.\frac{x^2+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)^2}{\left(x-1\right)^3}\)
\(=\frac{1}{x-1}\)
c) Với \(\forall x\)(\(x\ne1\)) thì biểu thức được xác định .
P/s : Theo mik câu c nên chuyển thành : Tìm x để biểu thức đạt giá trị nguyên.
Tại thấy câu c k khác j câu a !
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
diều kiện xác định là các mẫu phải khác o; số chia cũng khác o nhé:
ĐK: +) \(x+5\ne0\Rightarrow x\ne-5\)
+) \(2x-15\ne0\Rightarrow x\ne\frac{15}{2}\)
+) \(x^2-25\ne0\Rightarrow\left(x+5\right)\left(x-5\right)\ne0\Rightarrow x\ne\pm5\)
+) \(1-x\ne0\Rightarrow x\ne1\)
Vậy điều kiện xác đinh của A là : \(x\ne1;x\ne\frac{15}{2};x\ne\pm5\)