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Theo đề bài ta có :
\(\frac{x\left(3-x\right)}{x+1}\cdot\left(x+\frac{\left(3-x\right)}{x+1}\right)=2\)
=> \(\frac{\left(3x-x^2\right)}{x+1}\cdot\frac{\left(3-x+x^2+x\right)}{x+1}=2\)
=> \(\left(3x-x^2\right)\left(x^2+3\right)=2\left(x+1\right)^2\)
=> \(3x^3+9x-x^4-3x^2=2x^2+4x+2\)
=> \(3x^3+\left(9x-4x\right)+\left(-3x^2-2x^2\right)-x^4-2=0\)
=> \(3x^3+5x-5x^2-x^4-2=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x^3-1\right)=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x-1\right)\left(x^2+x+1\right)=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)-2\left(1-x\right)\left(x^2+x+1\right)=0\)
=> \(\left(1-x\right)\left(5x+x^3-2x^2-2x-2\right)=0\)
=> \(\left(1-x\right)\left(3x+x^3-2x^2-2\right)=0\)
=> \(\left(1-x\right)\left(x^3-x^2-x^2+x+2x-2\right)=0\)
=> \(\left(1-x\right)\left(x^2\left(x-1\right)-x\left(x-1\right)+2\left(x-1\right)\right)=0\)
=> \(\left(1-x\right)\left(x-1\right)\left(x^2-x+2\right)=0\)
Ta Thấy :
\(\left(x^2-x+2\right)=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\)
=> \(\hept{\begin{cases}1-x=0\\x-1=0\end{cases}}\)
=> x = 1
a. => 3-x2+x2-9=0
=> 3-9=0
=> -6=0 (vô lí)
Vạy ko có x thỏa mãn.
b. => x(x2-1/4)=0
=> x(x-1/2)(x+1/2)=0
=> x=0 hoặc x=1/2 hoặc x=-1/2
c. => x2(x-3)+4(3-x)=0
=> x2(x-3)-4(x-3)=0
=> (x-3)(x2-4)=0
=> (x-3)(x-2)(x+2)=0
=> x=3 hoặc x=2 hoặc x=-2
d. => [(2x-1)-(x+3)].[(2x-1)+(x+3)]=0
=> (2x-1-x-3)(2x-1+x+3)=0
=> (x-4)(3x+2)=0
=> x=4 hoặc 3x+2=0
=> x=4 hoặc x=-2/3.
áp dụng cosi a^2+1>=2a tương tự và cộng vế tương ứng suy ra đpcm
\(a^2+b^2+2\ge2\left(a+b\right)\)
\(\Leftrightarrow a^2+b^2+2-2\left(a+b\right)\ge0\)
\(\Leftrightarrow a^2+b^2+2-2a-2b\ge0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2\ge0\)( luôn đúng )
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}b-1=0\\b-1=0\end{cases}}\)\(\Leftrightarrow a=b=1\)
Vậy ...
a)x3-13x=0
<=>x(x2-13)=0
<=>x=0 hoặc x2-13=0<=>x2=13<=>x=\(^+_-\sqrt{13}\)
b)2-25x2=0
<=>25x2=2
<=>x2=2/25
<=>x=\(^+_-\sqrt{\frac{2}{25}}\)
c)x2=x+1/4
<=>4x2=4x+1
<=>4x2-4x-1=0
<=>(4x2-4x+1)-2=0
<=>(2x-1)2=2
*)2x-1=\(\sqrt{2}\)
<=>2x=\(\sqrt{2}\)+1
<=>x=(\(\sqrt{2}\)+1)/2
*)2x-1=-\(\sqrt{2}\)
<=>2x=-\(\sqrt{2}\)+1
<=>x=(-\(\sqrt{2}\)+1)/2
d)(2x-1)2=(x+3)2
<=>(2x-1)2-(x+3)2=0
<=>(2x-1-x-3)(2x-1+x+3)=0
<=>(x-4)(3x+2)=0
<=>x-4=0 hoặc 3x+2=0
<=>x=4 hoặc x=-2/3
a) 4x2 - 12x + 9 = 0 <=> (2x - 3)2 = 0 <=> 2x - 3 = 0 <=> x = 3/2.KL
b) ( 5 - 2x )( 2x + 7 ) + ( 25 - 4x2 ) = 0 <=> ( 5 - 2x )( 2x + 7 ) + ( 5 + 2x )( 5 - 2x ) = 0 <=> ( 5 - 2x )( 2x + 7 + 5 + 2x ) = 0. KL
<=> ( 5 - 2x )( 4x + 12 ) = 0 <=>\(\orbr{\begin{cases}5-2x=0\\4x+12=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=2\frac{1}{2}\\x=-3\end{cases}}\)KL.
c) ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 3 ) = 0 <=> ( x + 3 )( x2 - 3x + 9 + x - 3 ) = 0 <=> ( x + 3 )( x2 -2x + 6 ) = 0 <=> x + 3 = 0 (vi x2 - 2x + 6 = ( x + 1 )2 + 5 > 0 voi moi x) KL
<=>x=-3.KL
d) [ 2 ( 2x + 7 ) ]2 - [ 3 ( x + 3 ) ]2 = 0 <=> ( 4x + 14 )2 - ( 3x + 9 )2 = 0 <=> ( 4x + 14 + 3x + 9 )( 4x + 14 - 3x -9 ) = 0
<=> ( 7x + 23 )( x + 5 ) = 0 <=> 7x + 23 = 0 hoac x + 5 = 0 <=> x = -23/7 hoac x = -5.KL
a) <=> 3x-2=0 hoặc 4x+5=0
1) 3x-2=0 <=> 3x=2 <=> x=2/3
2) 4x+5=0 <=> 4x=-5 <=> x= -5/4
Bài 3:
Ta có:
\(81^8-1=\left(9^2\right)^8-1=\left[\left(3^2\right)^2\right]^8-1=3^{32}-1\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
Do đó:
\(A=3^4-1=80\)
\(5,4x^2-36=0\\ \Leftrightarrow\left(2x\right)^2-6^2=0\\ \Leftrightarrow\left(2x-6\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{3;-3\right\}\)
\(7,\left(3x+1\right)^2-16=0\\ \Leftrightarrow\left(3x+1\right)^2-4^2=0\\ \Leftrightarrow\left(3x+1-4\right)\left(3x+1+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-3=0\\3x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(S=\left\{1;-\dfrac{5}{3}\right\}\)
\(8,\left(2x-3\right)^2-49=0\\ \Leftrightarrow\left(2x-3\right)^2-7^2=0\\ \Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-10=0\\2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-2;5\right\}\)
Câu 6 đâu ạ?