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Mk xin lỗi nha, câu c sai đề
c) (x+6)4 + (x+8)4 = 272
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) 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)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)
\(\Leftrightarrow-x>-\dfrac{21}{2}\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
a, <=> (x-2)2=25
<=>x-2=5 hoặc x-2=-5
<=>x=7 hoặc x=-3
c,<=>(x2)2-16=0
<=>(x2)2=16
<=>x2=4
<=>x=2 hoặc x=-2
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.
\(\left(2x-1\right)^3+6\left(3x-1\right)^3=2\left(x+1\right)^3+6\left(x+2\right)^3\)
\(\Leftrightarrow\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1+1^3+6.\left[\left(3x\right)^3-3.\left(3x\right)^2.1+3.3x.1+1^3\right]=2\left(x^3+3x^2+3x+1\right)+6\left(x^2+3.x^2.2+3.x.2^2+2^3\right)\)
a) \(4.\left(x-1\right)^2-9=0\)
\(\Rightarrow4.\left(x-1\right)^2=9\)
\(\Rightarrow\left(x-1\right)^2=9:4=\dfrac{9}{4}=\left(\pm\dfrac{3}{2}\right)^2\)
\(\Rightarrow x-1=\pm\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
vậy\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
b) \(\dfrac{1}{4}-9.\left(x-1\right)^2=0\)
\(\Rightarrow9.\left(x-1\right)^2=\dfrac{1}{4}\)
\(\Rightarrow\left(x-1^2\right)=\dfrac{1}{36}=(\pm\dfrac{1}{6})^2\)
\(\Rightarrow x-1=\pm\dfrac{1}{6}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{6}\\x-1=\dfrac{-1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{5}{6}\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{5}{6}\end{matrix}\right.\)
e) \(\dfrac{1}{16}-\left(2x+\dfrac{3}{4}\right)^2=0\)
\(\Rightarrow\left(2x+\dfrac{3}{4}\right)^2=\dfrac{1}{16}=\left(\pm\dfrac{1}{4}\right)^2\)
\(\Rightarrow2x+\dfrac{3}{4}=\pm\dfrac{1}{4}\)
\(\Rightarrow\)\(\left[{}\begin{matrix}2x+\dfrac{3}{4}=\dfrac{1}{4}\\2x+\dfrac{3}{4}=\dfrac{-1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
a)
\(x^2-4x+4=25\)
\(\Leftrightarrow x^2-4x-21=0\)
\(\Leftrightarrow x^2+3x-7x-21=0\)
\(\Leftrightarrow x\left(x+3\right)-7\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b)
\(\dfrac{x-17}{1990}+\dfrac{x-21}{1986}+\dfrac{x+1}{1004}=4\)
\(\Leftrightarrow\dfrac{x-17}{1990}-1+\dfrac{x-21}{1986}-1+\dfrac{x+1}{1004}-2=4-1-1-2\)
\(\Leftrightarrow\dfrac{x-17-1990}{1990}+\dfrac{x-21-1986}{1986}+\dfrac{x+1-2008}{1004}=0\)
\(\Leftrightarrow\dfrac{x-2007}{1990}+\dfrac{x-2007}{1986}+\dfrac{x-2007}{1004}=0\)
\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\right)=0\)
\(\Leftrightarrow x-2007=0\) ( Vì: \(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\ne0\))
\(\Leftrightarrow x=2007\)
c.
\(4^x-12.2^x+32=0\)
\(\Leftrightarrow\left(2^x\right)^2-12.2^x+36-4=0\)
\(\Leftrightarrow2^x-2.2^x.6+6^2-2^2=0\)
\(\Leftrightarrow\left(2^x-6\right)^2-2^2=0\)
\(\Leftrightarrow\left(2^x-6-2\right)\left(2^x-6+2\right)=0\)
\(\Leftrightarrow\left(2^x-8\right)\left(2^x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2^x-8=0\\2^x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2^x=8\\2^x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)