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\(a,\)\(đkxđ\)của \(A\)\(:\)\(\hept{\begin{cases}x^2-25\ne0\\x^2+5x\ne0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)\left(x+5\right)\ne0\\x\left(x+5\right)\ne0\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x\ne\pm5\\x\ne0\end{cases}}\)
\(đkxđ\)của \(B\)\(:\)\(\hept{\begin{cases}x^2+5x\ne0\\5-x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\left(x+5\right)\ne0\\5-x\ne0\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x\ne\pm5\\x\ne0\end{cases}}\)
\(b,\)\(A=\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}=\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\)
\(=\frac{x^2-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}=\frac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\)\(=\frac{10x-25}{x\left(x+5\right)\left(x-5\right)}\)
\(B=\frac{2x-5}{x^2+5x}+\frac{x+3}{5-x}=\frac{2x-5}{x\left(x+5\right)}-\frac{x+3}{x-5}\)
\(=\frac{\left(2x-5\right)\left(x+5\right)-\left(x-3\right)\left(x^2+5x\right)}{x\left(x-5\right)\left(x+5\right)}\)
\(=\frac{2x^2+5x-25-x^3-2x^2+15x}{x\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-x^3+20x-25}{x\left(x-5\right)\left(x+5\right)}\)
\(\Rightarrow P=A:B=\frac{10x-25}{x\left(x+5\right)\left(x-5\right)}:\frac{x^3+20x-25}{x\left(x+5\right)\left(x-5\right)}\)
\(=\frac{10x-25}{x^3+20x-25}\)
Đề có vấn đề ko vậy babe -.- \(x^3+20x-25\)vẫn phân tích được, nhưng ko rút gọn được -.-
\(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)
\(\Leftrightarrow\frac{x^2+5}{\left(5-x\right)\left(5+x\right)}=\frac{3}{5+x}-\frac{x}{5-x}\)
\(\Leftrightarrow\frac{x^2+5}{\left(5-x\right)\left(5+x\right)}=\frac{3\left(5-x\right)-x\left(5+x\right)}{\left(5-x\right)\left(5+x\right)}\)
\(\Rightarrow x^2+5=3\left(5-x\right)-x\left(5+x\right)\)
\(\Leftrightarrow x^2+5=15-3x-5x-x^2\)
\(\Leftrightarrow15-3x-5x-x^2-x^2-5=0\)
\(\Leftrightarrow10-8x-2x^2=0\)
\(\Leftrightarrow2x^2+8x-10=0\)
\(\Leftrightarrow2\left(x^2+4x-5\right)=0\)
\(\Leftrightarrow2\left(x^2+5x-x-5\right)=0\)
\(\Leftrightarrow x^2-x+5x-5=0\)
\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}}\)
ĐK: ...
c) \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=\frac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)
\(\Leftrightarrow5x+25=0\)
\(\Leftrightarrow x=-5\)( ko t/m )
d) tương tự, ngại tính lắm
e) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1}{x^3-1}-\frac{3x^2}{x^3-1}=\frac{2x\left(x-1\right)}{x^3-1}\)
\(\Leftrightarrow4x^2-3x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=\frac{-1}{4}\left(c\right)\end{matrix}\right.\)
\(a.\frac{7x-3}{x-1}=\frac{3}{2}\)
\(\Leftrightarrow\frac{7x-3}{x-1}-\frac{3}{2}=0\)
\(\Leftrightarrow\frac{2\left(7x-3\right)}{2.\left(x-1\right)}-\frac{3\left(x-1\right)}{2\left(x-1\right)}=0\)
\(\Leftrightarrow\frac{14x-6-3x+3}{2\left(x-1\right)}=0\)
\(\Leftrightarrow11x-3=0\)
\(\Leftrightarrow x=\frac{3}{11}\)
\(b.\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)
\(\Leftrightarrow\frac{6-14x}{1+x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{2\left(6-14x\right)}{2\left(1+x\right)}-\frac{1+x}{2\left(1+x\right)}=0\)
\(\Leftrightarrow\frac{12-28x-1-x}{2\left(1+x\right)}=0\)
\(\Leftrightarrow11-29x=0\)
\(\Leftrightarrow x=\frac{11}{29}\)
\(c.\frac{1}{x-2}+3=\frac{3-x}{x-2}\)
\(\Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}-\frac{3-x}{x-2}=0\)
\(\Leftrightarrow\frac{1+3x-6-3+x}{x-2}=0\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow x=2\)
\(d.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
\(\Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}-\frac{20}{x^2-25}=0\)
\(\Leftrightarrow\frac{x^2+10x+25-x^2+10x-25-20}{x^2-25}=0\)
\(\Leftrightarrow20x-20=0\)
\(\Leftrightarrow x=10\)
a, \(3x-5=13\Leftrightarrow3x=18\Leftrightarrow x=6\)
b, \(4x-2=3x+1\Leftrightarrow x=3\)
c, \(5\left(x-3\right)-2\left(x-5\right)=58\Leftrightarrow5x-15-2x+10=58\)
\(\Leftrightarrow3x-5=58\Leftrightarrow3x=63\Leftrightarrow x=21\)
d, \(mx+5x=m^2m^2-25\Leftrightarrow x\left(m+5\right)=m^4-25\)
Bài 2 Tìm x biết
a) 3x -5 = 13
<=> 3x = 18
<=> x = 6
Vậy x = 6
b) 4x - 2 = 3x + 1
<=> 4x - 3x = 2 + 1
<=> x = 3
Vậy x = 3
c) 5(x - 3) - 2(x - 5) = 58
<=> 5x - 15 - 2x + 10 = 58
<=> 3x - 5 = 58
<=> 3x = 63
<=> x = 21
Vậy x = 21
d) mx + 5x = m2 - 25
<=> mx + 5x + 25 - m2 = 0
<=> x(5 + m) + (5 - m)(5 + m) = 0
<=> (5 + m)(x + 5 - m) = 0
<=> \(\left[{}\begin{matrix}5+m=0\\x+5-m=0\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}m=-5\\x+5-m=0\end{matrix}\right.\) => x + 5 - (-5) = 0
<=> x + 10 = 0
<=> x = -10
Vậy x = -10
#Không chắc lắm :)
Bài 1:
a) \(\dfrac{3x^2-5}{x^2-5x}+\dfrac{5-15x}{5x-25}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{5\left(1-3x\right)}{5\left(x-5\right)}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{1-3x}{x-5}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{x\left(1-3x\right)}{x\left(x-5\right)}\)
\(=\dfrac{3x^2-5+x\left(1-3x\right)}{x\left(x-5\right)}\)
\(=\dfrac{3x^2-5+x-3x^2}{x\left(x-5\right)}\)
\(=\dfrac{-5+x}{x\left(x-5\right)}\)
\(=\dfrac{x-5}{x\left(x-5\right)}\)
\(=\dfrac{1}{x}\)
b) \(\dfrac{4+x^3}{x-3}-\dfrac{2x+2x^2}{x-3}+\dfrac{2x-13}{x-3}\)
\(=\dfrac{\left(4+x^3\right)-\left(2x+2x^2\right)+\left(2x-13\right)}{x-3}\)
\(=\dfrac{4+x^3-2x-2x^2+2x-13}{x-3}\)
\(=\dfrac{x^3-2x^2-9}{x-3}\)
\(=\dfrac{x^3-3x^2+x^2-9}{x-3}\)
\(=\dfrac{x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)}{x-3}\)
\(=\dfrac{\left(x-3\right)\left(x^2+x+3\right)}{x-3}\)
\(=x^2+x+3\)
c) \(\dfrac{2}{x-5}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2\left(x+5\right)+x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2x+10+x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3}{x+5}\)
d) Đề sai?
Bài 2:
\(A=2\left(x+1\right)+\left(3x+2\right)\left(3x-2\right)-9x^2\)
\(A=2x+2+9x^2-4-9x^2\)
\(A=2x-2\)
\(A=2\left(x-1\right)\)
Thay x = 15 vào A ta được:
\(A=2\left(15-1\right)\)
\(A=2.14=28\)
d: \(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8=12x^2-12x-8\)
\(\Leftrightarrow12x^2+16=12x^2-12x-8\)
=>-12x=24
hay x=-2
e: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(x-5\right)^2\)
\(\Leftrightarrow x^2+7x+10-12x+9=x^2-10x+25\)
=>-5x+19=-10x+25
=>5x=6
hay x=6/5
f: \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
=>x-105=0
hay x=105
Ta có
C = ( x + 5 ) 2 + ( x - 5 ) 2 ( x 2 + 25 ) = x 2 + 2 . x . 5 + 5 2 + x 2 - 2 . x . 5 + 5 2 ( x 2 + 25 ) = x 2 + 10 x + 25 + x 2 - 10 x + 25 x 2 + 25 = 2 ( x 2 + 25 ) x 2 + 25 = 2
D = ( 2 x + 5 ) 2 + ( 5 x - 2 ) 2 x 2 + 1 = 4 x 2 + 2 . 2 x . 5 + 5 2 + 25 x 2 - 2 . 5 x . 2 + 2 2 x 2 + 1 = 29 x 2 + 29 x 2 + 1 = 29 ( x 2 + 1 ) x 2 + 1 = 29
Vậy D = 29; C = 2 suy ra D = 14C + 1 (do 29 = 14.2 + 1)
Đáp án cần chọn là: A