Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1/
a/ \(x^2+\left(y-10\right)^2=0\)
vì: \(\left\{{}\begin{matrix}x^2\ge0\forall x\\\left(y-10\right)^4\ge0\forall y\end{matrix}\right.\)
=> Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y-10=0\Rightarrow y=10\end{matrix}\right.\)
vậy......
b/ \(\left(0,5x-5\right)^{20}+\left(y^2-0,25\right)^{10}\le0\)
vì: \(\left\{{}\begin{matrix}\left(0,5x-5\right)^{20}\ge0\forall x\\\left(y^2-0,25\right)^2\ge0\forall y\end{matrix}\right.\)=> \(\left(0,5x-5\right)^{20}+\left(y^2-0,25\right)^{10}\ge0\)
=> Dấu ''='' xảy ra khi :
\(\left\{{}\begin{matrix}0,5x-5=0\\y^2-0,25=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{0,5}=10\\y^2=0,25\Rightarrow\left[{}\begin{matrix}y=0,5\\y=-0,5\end{matrix}\right.\end{matrix}\right.\)
Vậy........
2/ Ta có: \(2011\equiv1\left(mod10\right)\)
\(2011^{201}\equiv1^{201}\equiv1\left(mod10\right)\);
Có: \(1997^3\equiv3\left(mod10\right)\)
\(\left(1997^3\right)^4\equiv3^4\equiv1\left(mod10\right)\)
\(\left(1997^{12}\right)^{14}\equiv1^{14}\equiv1\left(mod10\right)\) hay \(1997^{168}\equiv1\left(mod10\right)\)
=> \(2011^{201}-1997^{168}\equiv1-1\equiv0\left(mod10\right)\)
hay \(2011^{201}-1997^{168}\) chia hết cho 10
=> Đpcm
1,
\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)
2,
\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)
3, Làm tương tự câu 2
5,
\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)
6,
\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)
7,
\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)
9,
a) Ta có:
\(\left|x-2017\right|\ge0\) với \(\forall x\)
\(\left|y-2018\right|\ge0\) với \(\forall x\)
\(\Rightarrow\left|x-2017\right|+\left|y-2018\right|\ge0\) với \(\forall x\)
\(\Rightarrow\) Không có giá trị của x; y thỏa mãn yêu cầu
Vậy \(x;y\in\varnothing\)
b) Ta có:
\(3.\left|x-y\right|^5\ge0\)
\(10.\left|y+\dfrac{2}{3}\right|^7\ge0\)
\(3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7\ge0\left(1\right)\)
Theo bài ra ta có: \(3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7\le0\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7=0\)
\(\Rightarrow\left\{{}\begin{matrix}3.\left|x-y\right|^5=0\\10.\left|y+\dfrac{2}{3}\right|^7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-y\right|^5=0\\\left|y+\dfrac{2}{3}\right|^7=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x-y=0\\y+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=y\\y=\dfrac{-2}{3}\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\y=\dfrac{-2}{3}\end{matrix}\right.\)\(\)
\(\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}\Leftrightarrow\dfrac{2x^2}{18}=\dfrac{y^2}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x^2}{18}=\dfrac{y^2}{16}=\dfrac{2x^2+y^2}{18+16}=\dfrac{136}{34}=4\)
Suy ra: \(\left\{{}\begin{matrix}x^2=4.9=36\\y^2=4.16=64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm8\end{matrix}\right.\)
2) Ta có: \(2^{20}=\left(2^4\right)^5=16^5\)
Được biết số có tận cùng là \(6\) thì lũy thừa bao nhiêu cũng bằng \(6\)
Nên \(16^5=\overline{...6}\Leftrightarrow16^5-1=\overline{.....5}⋮5\)
Nên \(\dfrac{2^{20}-1}{5}\) là số nguyên
3)
Ta có:
\(A=100^2+200^2+...+1000^2\)
\(A=\left(1.100\right)^2+\left(2.100\right)^2+...+\left(10.100\right)^2\)
\(A=1^2.100^2+2^2.100^2+....+10^2.100^2\)
\(A=100^2\left(1^2+2^2+...+100^2\right)\)
\(A=10000.385=3850000\)
Bài 1:
a: \(=\dfrac{-1}{8}+1-\dfrac{9}{4}-1\)
\(=\dfrac{-1}{8}-\dfrac{18}{8}=\dfrac{-19}{8}\)
b: \(=4\cdot1-2\cdot\dfrac{1}{4}+3\cdot\dfrac{-1}{2}+1\)
\(=4-\dfrac{1}{2}-\dfrac{3}{2}+1\)
=5-2
=3
I . Trắc Nghiệm
1B . 2D . 3C . 5A
II . Tự luận
2,a,Ta có: A+(x\(^2\)y-2xy\(^2\)+5xy+1)=-2x\(^2\)y+xy\(^2\)-xy-1
\(\Leftrightarrow\) A=(-2x\(^2\)y+xy\(^2\)-xy-1) - (x\(^2\)y-2xy\(^2\)+5xy+1)
=-2x\(^2\)y+xy\(^2\)-xy-1 - x\(^2\)y+2xy\(^2\)-5xy-1
=(-2x\(^2\)y - x\(^2\)y) + (xy\(^2\)+ 2xy\(^2\)) + (-xy - 5xy ) + (-1 - 1)
= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
b, thay x=1,y=2 vào đa thức A
Ta có A= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
= -3 . 1\(^2\) . 2 + 3 .1 . 2\(^2\) - 6 . 1 . 2 -2
= -6 + 12 - 12 - 2
= -8
3,Sắp xếp
f(x) =9-x\(^5\)+4x-2x\(^3\)+x\(^2\)-7x\(^4\)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x
g(x) = x\(^5\)-9+2x\(^2\)+7x\(^4\)+2x\(^3\)-3x
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
b,f(x) + g(x)=(9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x) + (-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
=(9-9)+(-x\(^5\)+x\(^5\))+(-7x\(^4\)+7x\(^4\))+(-2x\(^3\)+2x\(^3\))+(x\(^2\)+2x\(^2\))+(4x-3x)
= 3x\(^2\) + x
g(x)-f(x)=(-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x) - (9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x)
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x-9+x\(^5\)+7x\(^4\)+2x \(^3\)-x\(^2\)-4x
=(-9-9)+(x\(^5\)+x\(^5\))+(7x\(^4\)+7x\(^4\))+(2x\(^3\)+2x\(^3\))+(2x\(^2\)-x\(^2\))+(3x-4x)
= -18 + 2x\(^5\) + 14x\(^4\) + 4x\(^3\) + x\(^2\) - x
a: Ta có: \(x^2\ge0\forall x\)
\(\left(y-\dfrac{1}{10}\right)^4\ge0\forall y\)
Do đó: \(x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(0;\dfrac{1}{10}\right)\)