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Mk xin lỗi nha, câu c sai đề
c) (x+6)4 + (x+8)4 = 272
a) ĐKXĐ: x khác 0
\(x+\dfrac{5}{x}>0\)
\(\Leftrightarrow x^2+5>0\) ( luôn đúng)
Vậy bất pt vô số nghiệm ( loại x = 0)
d)
\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)
\(\Leftrightarrow2x+2-4x+4>-15\)
\(\Leftrightarrow-2x>-21\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
Vậy....................
a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)
Mà \(x^2+5>0\)
\(\Rightarrow x>0\)
d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)
\(\Leftrightarrow-x>-\dfrac{21}{2}\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
a.
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow\left(x^2+5x\right)-\left(2x+10\right)=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b.
\(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=1\end{matrix}\right.\)
bài 2:
ĐKXĐ: x khác -1
\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\)
\(\Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x+1}\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow0x=-1\)
\(\Leftrightarrow x\in\varnothing\)
Suy ra pt vô nghiệm
b.
ĐKXĐ: x khác \(\dfrac{3}{2}\)
\(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow\dfrac{x^2+4x+4}{2x-3}-\dfrac{2x-3}{2x-3}=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\) ( loại)
a) (x + 5)2 - (x - 3)2 = 2x - 7
(x + 5 - x + 3)(x + 5 + x - 3) = 2x - 7
8(2x + 2)= 2x - 7
16x + 16 = 2x - 7
16x - 2x = - 7 - 16
14x = - 23
x = - 23/14
b) (2x - 3)(4x2 + 6x + 9) = 98
(2x)3 - 33 = 98
8x3 - 27 = 98
8x3 = 125
x3 = 125/8
x3 = (5/2)3
x = 5/2
a: \(\Leftrightarrow5\left(x+1\right)\left(x-1\right)=2x-2-3x-3=-x-5\)
\(\Leftrightarrow5x^2-5+x+5=0\)
=>x(5x+1)=0
=>x=0 hoặc x=-1/5
b: \(\Leftrightarrow x^2-x-\left(2x-3\right)\left(x+1\right)=2x+3\)
\(\Leftrightarrow x^2-x-2x^2-2x+3x+3=2x+3\)
\(\Leftrightarrow-x^2+3=2x+3\)
=>-x(x+2)=0
=>x=0(nhận) hoặc x=-2(nhận)
c: \(\Leftrightarrow4x^2-25=0\)
=>(2x-5)(2x+5)=0
=>x=5/2 hoặc x=-5/2
2: \(\Leftrightarrow\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)=2\left(x-1\right)\left(x+1\right)\)
=>x^2-3x-4+x^2+3x-4=2x^2-2
=>2x^2-8=2x^2-2(loại)
3: \(\Leftrightarrow\left(x^2-x\right)\left(x-3\right)+x^2\left(x+3\right)=-7x^2+3x\)
=>x^3-3x^2-x^2+3x+x^3+3x^2+7x^2-3x=0
=>2x^3+6x^2=0
=>2x^2(x+3)=0
=>x=0(nhận) hoặc x=-3(loại)
1: =>x^3-5x^2+x^2-5x+3x-15=0
=>(x-5)(x^2+x+3)=0
=>x-5=0
=>x=5
2: =>x^3+6x^2+12x+35=0
=>x^3+5x^2+x^2+5x+7x+35=0
=>(x+5)(x^2+x+7)=0
=>x+5=0
=>x=-5
3: \(\Leftrightarrow\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}+1\right)\)
=>x+100=0
=>x=-100