Tìm x,y biết
a,\(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=994-15:3+1^{2025}\)
b,\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
c,\(2024^{|x-1|+y^2-1}\cdot3^{2024}=9^{1012}\)
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a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
=>\(8\cdot x+1\cdot x=3305+1\)
=>\(9x=3306\)
=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)
b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)
=>\(2^x\left(1+2+4+8\right)=480\)
=>\(2^x\cdot15=480\)
=>\(2^x=32\)
=>\(2^x=2^5\)
=>x+5
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
\(\left(x+1\right)\left(y-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0-1\\y=0+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)
Vậy x = - 1 ; y = 2
a) 15 - 3x = 6
=> 3x = 15 - 6
=> 3x = 9
=> x = 9 : 3 = 3
b) \(\dfrac{2}{3}-\dfrac{4}{3}x=\dfrac{1}{2}\Rightarrow\dfrac{4}{3}x=\dfrac{2}{3}-\dfrac{1}{2}\Rightarrow\dfrac{4}{3}x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{6}:\dfrac{4}{3}\Rightarrow x=\dfrac{3}{24}=\dfrac{1}{8}\)c) 319.(x - 12 ) = 4 . 320
=> x - 12 = 4 . 3
=> x - 12 = 12
=> x = 12 + 12 = 24
d) 3.(x - 4) = 2^2 . 3^3
=> x - 4 = 2^2 . 3^2
=> x - 4 = 36
=> x = 36 + 4 = 40
a)\(15-3x=6\Rightarrow3x=9\Rightarrow x=3\)
b) \(\dfrac{2}{3}-\dfrac{4}{3}x=\dfrac{1}{2}\Rightarrow\dfrac{4}{3}x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{8}\)
c) \(3^{19}.\left(x-12\right)=4.3^{20}\Rightarrow x-12=12\Rightarrow x=24\)
d)\(3\left(x-4\right)=2^2.3^3\Rightarrow3x-12=108\Rightarrow3x=120\Rightarrow x=40\)
a)
( 4x - 9 ) ( 2,5 + (-7/3) . x ) = 0
\(\Rightarrow\orbr{\begin{cases}4x-9=0\\2,5+\frac{-7}{3}x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)
P/s: đợi xíu làm câu b
b) \(\frac{1}{x\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2015}\)
\(\frac{-1}{x+3}=\frac{1}{2015}\)
\(\Leftrightarrow x+3=-2015\)
\(\Leftrightarrow x=-2018\)
Vậy,.........
\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\dfrac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)
\(=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)
\(=\dfrac{3^{29}.2^3}{2^2.3^{28}}\)
\(=\dfrac{3.2}{1.1}=6\)
Lời giải:
Ta thấy, với mọi $x,y,z$ là số thực thì:
$(x-y+z)^2\geq 0$
$\sqrt{y^4}\geq 0$
$|1-z^3|\geq 0$
$\Rightarrow (x-y+z)^2+\sqrt{y^4}+|1-z^3|\geq 0$ với mọi $x,y,z$
Kết hợp $(x-y+z)^2+\sqrt{y^4}+|1-z^3|\leq 0$
$\Rightarrow (x-y+z)^2+\sqrt{y^4}+|1-z^3|=0$
Điều này xảy ra khi: $x-y+z=y^4=1-z^3=0$
$\Leftrightarrow y=0; z=1; x=-1$
a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\)
8 .x + 1 . x = 990
x . [ 8 +1 ] = 990
x . 9 = 990
x = 990 : 9
x = 110
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