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,\(2024^{|x-1|+y^2-1}\cdot3^{2024}=9^{1012}\)
Những câu hỏi liên quan
tìm x,y biết
a,\(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
b,\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
c,\(2024^{\left|x-1\right|=y^2-1}\cdot3^{2024}=9^{1012}\)
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
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Bài 1: Tínha. left(1+frac{1}{1cdot3}right)cdotleft(1+frac{1}{2cdot4}right)cdotleft(1+frac{1}{3cdot5}right)+left(1+frac{1}{4cdot6}right).....left(1+frac{1}{99cdot101}right)b. left[sqrt{0,64}+sqrt{0,0001}-sqrt{left(-0,5right)^2}right]divleft[3cdotsqrt{left(0,04right)^2}-sqrt{left(-2right)^4}right]c. frac{5.4^{15}cdot9^9-4.3^{20}cdot8^9}{5cdot2^9cdot6^{19}-7cdot2^{29}cdot27^6}-frac{2^{19}cdot6^{15}-7cdot6^{10}cdot2^{20}cdot3^6}{9cdot6^{19}cdot2^9-4cdot3^{17}cdot2^{26}}+0,left(6right)Bài 2: Tìm x, y...
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Bài 1: Tính
a. \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)+\left(1+\frac{1}{4\cdot6}\right).....\left(1+\frac{1}{99\cdot101}\right)\)
b. \(\left[\sqrt{0,64}+\sqrt{0,0001}-\sqrt{\left(-0,5\right)^2}\right]\div\left[3\cdot\sqrt{\left(0,04\right)^2}-\sqrt{\left(-2\right)^4}\right]\)
c. \(\frac{5.4^{15}\cdot9^9-4.3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot6^{15}-7\cdot6^{10}\cdot2^{20}\cdot3^6}{9\cdot6^{19}\cdot2^9-4\cdot3^{17}\cdot2^{26}}+0,\left(6\right)\)
Bài 2: Tìm x, y, z biết :
a. \(\left(x-10\right)^{1+x}=\left(x-10\right)^{x+2009}\left(x\in Z\right)\)
b. \(\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\left(x,y\in N\right)\)
c. \(25-y^2=8\left(x-2009\right)^2\left(x,y\in Z\right)\)
d. \(2008\left(x-4\right)^2+2009\left|x^2-16\right|+\left(y+1\right)^2\le0\)
e. \(2x=3y\) ; \(4z=5x\) và \(3y^2-z^2=-33\)
Bài 3: Chứng minh rằng
a. \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2009^2}>\frac{1}{2009}\)
b. \(\left[75\cdot\left(4^{2008}+4^{2007}+4^{2006}+...+4+1\right)+25\right]⋮100\)
Bài 4:
a. Tìm giá trị nhỏ nhất của biểu thức : \(M=\left(x^2+2\right)+\left|x+y-2009\right|+2005\)
b. So sánh: \(31^{11}\) và \(\left(-17\right)^{14}\)
c. So sánh: \(\left(\frac{9}{11}-0,81\right)^{2012}\) và \(\frac{1}{10^{4024}}\)
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
Tìm x , y \(\in Z\)biết :
a) \(\left(x+1\right)\cdot\left(y-2\right)=0\)
b) \(\left(x+4\right)\cdot\left(y-2\right)=2\)
c) \(x\cdot y+5\cdot x+y=4\)
d) \(3\cdot x+4\cdot y-x\cdot y=15\)
\(\left(x+1\right)\left(y-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0-1\\y=0+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)
Vậy x = - 1 ; y = 2
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Tìm x,y,z biết :a)frac{3}{4}cdotleft(x+frac{1}{2}right)frac{1}{2}b) xcdotleft(x-frac{3}{2}right)0c)|frac{1}{4}-x|+frac{2}{3}frac{1}{2}d)|3cdot x-2||x+1|e)|x+frac{3}{4}|+|x-y|0g)|4cdot x-2|-2cdot xfrac{3}{4}h)|5cdot x+1|-2cdot x1i)(x-frac{3}{2})cdotleft(2cdot x+1right)0k)|x+frac{1}{2}|+|x+y+z|+|frac{1}{3}+y|0l)|2cdot x-frac{1}{2}|-frac{1}{4}3GIÚP MK VS, CẢM ƠN MỌI NGƯỜI RẤT NHÌU !!!
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Tìm x,y,z biết :
a)\(\frac{3}{4}\cdot\left(x+\frac{1}{2}\right)=\frac{1}{2}\)
b) \(x\cdot\left(x-\frac{3}{2}\right)=0\)
c)\(|\frac{1}{4}-x|+\frac{2}{3}=\frac{1}{2}\)
d)\(|3\cdot x-2|=|x+1|\)
e)\(|x+\frac{3}{4}|+|x-y|=0\)
g)\(|4\cdot x-2|-2\cdot x=\frac{3}{4}\)
h)\(|5\cdot x+1|-2\cdot x=1\)
i)\((x-\frac{3}{2})\cdot\left(2\cdot x+1\right)>0\)
k)\(|x+\frac{1}{2}|+|x+y+z|+|\frac{1}{3}+y|=0\)
l)\(|2\cdot x-\frac{1}{2}|-\frac{1}{4}=3\)
GIÚP MK VS, CẢM ƠN MỌI NGƯỜI RẤT NHÌU !!!
lam sao bạn viết chữ to như 3/4 . ( x + 1/2 ) vậy
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Tìm x
a ) \(15-3\cdot x=6\) b)\(\dfrac{2}{3}-\dfrac{4}{3}\cdot x=\dfrac{1}{2}\) c) \(3^{19}\cdot\left(x-12\right)=4\cdot3^{20}\) d) \(3\cdot\left(x-4\right)=2^2\cdot3^3\)
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
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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)\(3\left(x-4\right)=2^2.3^3\Rightarrow3x-12=108\Rightarrow3x=120\Rightarrow x=40\)
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Tìm x biết :
a, ( 4x - 9 ) . ( 2,5 + \(\frac{-7}{3}\). x ) = 0
b, \(\frac{1}{x\cdot\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\cdot\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)
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
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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,.........
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A/ Ta có số nào nhân với 0 cx = 0
Vậy từ đó suy ra 2 trường hợp
TH1\(4x-9=0\)
\(=>x=\frac{9}{4}\)
TH2 \(2,5+-\frac{7}{3}x=0\)
\(=>x=\frac{15}{14}\)
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1:rút gọn
\(\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}\)
2: tìm x
\(\dfrac{3\cdot\left(x-2\right)}{4}-\dfrac{2\cdot\left(1+2x\right)}{3}=1\dfrac{1}{4}-5\cdot\dfrac{\left(1+3x\right)}{6}-\dfrac{x-2}{12}\)
\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\dfrac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)
\(=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)
\(=\dfrac{3^{29}.2^3}{2^2.3^{28}}\)
\(=\dfrac{3.2}{1.1}=6\)
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cho các số thực x,y,z thỏa mãn \(\left(x-y +z\right)^2\)+\(\sqrt{y^4}\)+\(\left|1-z^3\right|\) \(\le\) 0
Chứng minh rằng \(x^{2023}\)+\(y^{2024}\)+\(z^{2025}\)=0
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: $x-y+z=y^4=1-z^3=0$
$\Leftrightarrow y=0; z=1; x=-1$
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Giải phương trình
\(1,\dfrac{x^2-2x-3}{x-1}+\dfrac{x^2-8x+20}{x-4}=\dfrac{x^2-4x+6}{x-2}+\dfrac{x^2-6x+12}{x-3}\)
\(2,\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot[1+\dfrac{1}{x\cdot\left(x+2\right)}]=\dfrac{31}{16}\left(x\in N\right)\)
