x^64=x
(x-1)^4=9X-1)^2
giải hộ mình nhé
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\(\dfrac{3}{x}=\dfrac{1}{y}=\dfrac{6}{2}\)
\(\Rightarrow\dfrac{3}{x}=\dfrac{1}{y}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{3}\end{matrix}\right.\)
Ta có :
\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2-\frac{1}{64}=0\)
\(\Leftrightarrow\)\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2=\frac{1}{64}\)
\(\Leftrightarrow\)\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2=\frac{1^2}{8^2}\)
\(\Leftrightarrow\)\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2=\left(\frac{1}{8}\right)^2\)
\(\Leftrightarrow\)\(\frac{3}{4}x-\frac{1}{2}=\frac{1}{8}\)
\(\Leftrightarrow\)\(\frac{3}{4}x=\frac{1}{8}+\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{3}{4}x=\frac{5}{8}\)
\(\Leftrightarrow\)\(x=\frac{5}{8}:\frac{3}{4}\)
\(\Leftrightarrow\)\(x=\frac{5}{8}.\frac{4}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{2}.\frac{1}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}\)
Chúc bạn học tốt ~
\(\left(\frac{3}{4}.x-\frac{1}{2}\right)^2-\frac{1}{64}=0\)
\(\left(\frac{3}{4}.x-\frac{1}{2}\right)^2=0+\frac{1}{64}=\frac{1}{64}\)
\(\left(\frac{3}{4}.x-\frac{1}{2}\right)^2=\left(\frac{1}{8}\right)^2\)
=>\(\frac{3}{4}.x-\frac{1}{2}=\frac{1}{8}\)
\(\frac{3}{4}.x=\frac{1}{8}+\frac{1}{2}\)
\(\frac{3}{4}.x=\frac{5}{8}\)
\(x=\frac{5}{8}:\frac{3}{4}\)
\(x=\frac{5}{6}\)
Mấy cái bước suy ra ≥;≤ là có công thức hay là định lý gì không ạ ?
\(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)=x^2y-xy^2+y^2z-yz^2+z^2z-zx^2=x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(z-y\right)\)
\(x^2\left(y-z\right)-y^2\left(x-z\right)-z^2\left(y-z\right)=\left(y-z\right)\left(x-z\right)\left(x+z\right)-y^2\left(x-z\right)=\left(x-z\right)\left(xy-yz-zx-z^2-y^2\right)\)
t cx k bt có đúng hay k đâu nha, nhớ xem kĩ lại
1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
Câu 1:
0,9 x 218 x 2 + 0,18 x 4290 + 0,6 x 353 x 3
= 9/10 x 436 + 9/50 x 4290 + 6/10 x 1059
= 9 x 43,6 + 9 x 85,8 + 6 x 105,9
= 3 x 130,8 + 3 x 257,4 + 3 x 211,8
= 3 x ( 130,8 + 257,4 + 211,8 )
= 3 x 600
= 1800
Câu 2:
3/4 x X + 1/2 x X - 15 = 35
X x ( 3/4 + 1/2 ) - 15 = 35
X x ( 3/4 + 1/2 ) = 50
X x 5/4 = 50
X = 40
VẬy X = 40
đáp số cuối cùng là 0 vì 1-1/4 = 0/4 = 0 x cho những số nào khác cũng bằng 0 thôi
\(\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times\left(1-\frac{1}{16}\right)\times...\times\left(1-\frac{1}{10000}\right)\)
\(=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times...\times\frac{999}{10000}\)
\(=\frac{1\times3}{2\times2}\times\frac{2\times4}{3\times3}\times\frac{3\times5}{4\times4}\times...\times\frac{99\times101}{100\times100}\)
\(=\frac{1}{2}\times\frac{101}{100}\)
\(=\frac{101}{200}\)
\(x^{64}=x\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}\)
vậy \(\hept{\begin{cases}x=0\\x=1\end{cases}}\)
\(\left(x-1\right)^4=9\left(x-1\right)^2\)
\(\left(x-1\right)^4-9\left(x-1\right)^2=0\)
\(\left(x-1\right)^2.\left[\left(x-1\right)^2-3^2\right]=0\)
\(\left(x-1\right)^2.\left(x-1-3\right)\left(x-1+3\right)=0\)
\(\left(x-1\right)^2.\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow\left(x-1\right)^2=0\)hoặc \(\orbr{\begin{cases}x-4=0\\x-2=0\end{cases}}\)
\(\Rightarrow x-1=0\)hoặc \(\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
vậy \(x=1\)hoặc \(\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
x = -1;0;1.
x = 0 .