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TT
23 tháng 5 2021
a) \(A=x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+5\right)\)
\(A=2x^2+x-x^3-2x^2+x^3-x+5\)
\(A=5\)
=> giá trị biểu thức ko phụ thuộc vào biến x
b) \(A=x\left(3x^2-x+5\right)-\left(2x^3+3x-16\right)-x\left(x^2-x+2\right)\)
=> \(A=3x^3-x^2+5x-2x^3-3x+16-x^3+x^2-2x\)
=> \(A=\)16
vậy giá trị của biểu thức A ko phụ thuộc vào biến x
18 tháng 11 2022
a: \(\dfrac{7x^3y^4}{35xy}=\dfrac{7xy\cdot x^2y^3}{7xy\cdot5}=\dfrac{x^2y^3}{5}\)
b: \(\dfrac{x^3-4x}{10-5x}=\dfrac{-x\left(x-2\right)\left(x+2\right)}{5\left(x-2\right)}=\dfrac{-x\left(x+2\right)}{5}=\dfrac{-x^2-2x}{5}\)
c: \(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x+2}{x-1}\)
d: \(\left(x^2-x-2\right)\left(x-1\right)\)
\(=\left(x-2\right)\left(x+1\right)\left(x-1\right)\)
\(=\left(x^2-3x+2\right)\left(x+1\right)\)
=>\(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)
e: \(\dfrac{x^3+8}{x^2-2x+4}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\)
6 tháng 3 2022
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
CM
10 tháng 6 2019
Gọi T = (x – 1)(x – 2)(x + 4)(x + 5) – 27
= [(x – 1)(x + 4)].[(x – 2)(x + 5)] – 27
= ( x 2 + 3x – 4).( x 2 + 3x – 10) – 27
Đặt x 2 + 3x – 7 = t
Từ đó ta có T = (t – 3)(t + 3) – 27 = t 2 – 9 – 27 = t 2 – 36 = (t – 6)(t + 6)
Thay t = x 2 + 3x – 7 ta được
T = ( x 2 + 3x – 7 – 6)( x 2 + 3x – 7 + 6)
= ( x 2 + 3x – 13)( x 2 + 3x – 1) suy ra a = -13; b = -1 => a + b = -14
Đáp án cần chọn là: D
TA
2
1 tháng 3 2021
1) ĐKXĐ: \(x\notin\left\{1;-2\right\}\)
Ta có: \(\dfrac{2x}{x-1}-\dfrac{1}{x+2}=2\)
\(\Leftrightarrow\dfrac{2x\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+2\right)}=\dfrac{2\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)
Suy ra: \(2x^2+4x-x+1=2\left(x^2+x-2\right)\)
\(\Leftrightarrow2x^2+3x+1=2x^2+2x-4\)
\(\Leftrightarrow2x^2+3x+1-2x^2-2x+4=0\)
\(\Leftrightarrow x+5=0\)
hay x=-5(thỏa ĐK)
Vậy: S={-5}
2) ĐKXĐ: \(x\notin\left\{5;-5\right\}\)
Ta có: \(\dfrac{x}{x^2-25}-\dfrac{1-x}{x-5}=\dfrac{1}{x+5}\)
\(\Leftrightarrow\dfrac{x}{\left(x-5\right)\left(x+5\right)}+\dfrac{\left(x-1\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(x+x^2+5x-x-5=x-5\)
\(\Leftrightarrow x^2+5x-5-x+5=0\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
Vậy: S={0;-4}
1 tháng 3 2021
a/ ĐKXĐ : \(x\ne1;-2\)
\(\dfrac{2x}{x-1}-\dfrac{1}{x+2}=2\)
\(\Leftrightarrow\dfrac{2x\left(x+2\right)-\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=2\)
\(\Leftrightarrow2x^2+3x-x+1=2x^2+4x-2x-4\)
\(\Leftrightarrow2x+1=2x-4\)
\(\Leftrightarrow1=-4\left(loại\right)\)
Vậy...
b/ĐKXĐ : \(x\ne\pm5\)
\(\dfrac{x}{x^2-25}-\dfrac{1-x}{x-5}=\dfrac{1}{x+5}\)
\(\Leftrightarrow\dfrac{x}{\left(x-5\right)\left(x+5\right)}+\dfrac{\left(x-1\right)\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow x+x^2+5x-x-5=x-5\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy...
Bài làm:
a) \(\frac{x}{x+5}+\frac{5}{x+5}=\frac{x+5}{x+5}=1\left(x\ne-5\right)\)
b) \(\frac{x^2}{x-1}-\frac{2x-1}{x-1}=\frac{x^2-2x+1}{x-1}=\frac{\left(x-1\right)^2}{x-1}=x-1\left(x\ne1\right)\)
HT^^
a) \(\frac{x}{x+5}+\frac{5}{x+5}=\frac{x+5}{x+5}=1\)
b) \(\frac{x^2}{x-1}-\frac{2x-1}{x-1}=\frac{x^2-2x-1}{x-1}\)
Bài đây tính hẳn bạn??