Bài 1:Tìm x,y biết:
a.\((x-\frac{2}{5})^{2010}+(y+\frac{3}{7})^{468}\le0\)
b.\((x+0,7)^{84}+(y-6,3)^{262}\le0\)
c.\((x-5)^{88}+(x+y+3)^{496}\le0\)
Bài 2: Tìm số nguyên dương x,y biết:
\(2^x-2^y=224\)
Tìm x,y biết :
a) \(\left(x-\frac{2}{5}\right)^{2010}\)+\(\left(y+\frac{3}{7}\right)^{468}\)\(\le0\)
B) \(\left(x+0,7\right)^{84}\)+ \(\left(y-6,3\right)^{262}\)\(\le0\)
c) \(\left(x-5\right)^{88}\)+\(\left(x+y+3\right)^{468}\)\(\le0\)
Gợi ý: Các biểu thức mũ chẵn đều không âm.
\(a^{2n}+b^{2n}\le0\Leftrightarrow a^{2n}+b^{2n}=0\Leftrightarrow a=b=0\)
a,\(\left(x-\frac{2}{5}\right)^{2010}+\left(y+\frac{3}{7}\right)^{468}\)< \(0\)
Vì \(\left(x-\frac{2}{5}\right)^{2010}\);\(\left(y+\frac{3}{7}\right)^{468}\)đều > \(0\)
=> \(\left(x-\frac{2}{5}\right)^{2010}=0\)
\(\left(y+\frac{3}{7}\right)^{468}=0\)
=> \(\left(x-\frac{2}{5}\right)^{2010}=0^{2010}\)
\(\left(y+\frac{3}{7}\right)^{468}=0^{468}\)
=> \(x-\frac{2}{5}=0\)
\(y-\frac{3}{7}=0\)
=> \(x=\frac{2}{5}\)
\(y=\frac{3}{7}\)
Vậy \(x=\frac{2}{5}\)\(y=\frac{3}{7}\)
b,\(\left(x+0,7\right)^{84}+\left(y-6,3\right)^{262}\)< \(0\)
Vì \(\left(x+0,7\right)^{84}\);\(\left(y-6,3\right)^{262}\)đều > \(0\)
=>\(\left(x+0,7\right)^{84}\) = \(0\)
\(\left(y-6,3\right)^{262}\) = \(0\)
=> \(x+0,7=0\)
\(y-6,3=0\)
=> \(x=0,7\)
\(y=-6,3\)
Vậy \(x=0,7\)\(y=-6,3\)
Bài 1: Chứng minh rằng
a) \(^{ }2010^{100}+2010^{99}\)chia hết cho 2011
b)\(25^{25}+5^{49}-125^{16}\)chia hết cho 29
c) \(9^7+81^4-27^5\)chia hết cho 7
Bài 2: Tìm x,y biết
a)\(\left(x-\frac{2}{5}\right)^{2010}+\left(y+\frac{3}{7}\right)^{468}\le0\)
b)\(\left(x+0.7\right)^{84}+\left(y-6.3\right)^{262}\le0\)
c)\(\left(x-5\right)^{88}+\left(x+y+3\right)^{496}\le0\)
Tìm x,y,z thuộc Q
a, \(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z+2004|\)
b, \(|x+\frac{9}{2}|+|y+\frac{4}{3}|+|z+\frac{7}{2}|\le0\)
c,\(|x+\frac{3}{4}|+|y-\frac{1}{5}|+|x+y+z|=0\)
d, \(|x+\frac{3}{4}|+|y-\frac{2}{5}|+|z+\frac{1}{2}|\le0\)
Bài 1:Thực hiện phép tính
\(a,\frac{\left(2^3.5.7\right)\left(5^2.7^3\right)}{(2.5.7^2)^2}\)
\(b,\frac{4}{77}+\frac{4}{165}+\frac{4}{285}:\frac{5}{84}+\frac{3}{180}+\frac{4}{285}\)
Bài 2:Tìm x,y
\(a,\left(x+\frac{2}{3}\right).\frac{-3}{5}+\frac{4}{7}=1\frac{4}{7}.x\)
\(b,|5x-2|\le0\)
\(c,(x-7).(x+3)< 0\)
\(d,(x-3).(2y+1)=7\)\((với\)\(x,y\)nguyên)
a, \(\frac{\left(2^3.5.7\right)\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)\(=\frac{2^3.5.7.5^2.7^3}{2^2.5^2.7^4}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=10\)
b, \(\frac{4}{77}+\frac{4}{165}+\frac{4}{285}\)
\(=\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}\)
\(=\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}\)
\(=\frac{1}{7}-\frac{1}{19}\)
\(=\frac{19}{133}-\frac{7}{133}=\frac{12}{133}\)
Bài 2:
\(a,\left(x+\frac{2}{3}\right).\frac{-3}{5}+\frac{4}{7}=1\frac{4}{7}.x\)
\(\Rightarrow\frac{-3}{5}x+\frac{-2}{5}+\frac{4}{7}=\frac{11}{7}.y\)
\(\Rightarrow\frac{-3}{5}x+\frac{6}{35}=\frac{11}{7}.y\)
Từ đây làm nốt
b, \(\left|5x-2\right|\le0\)
\(\Rightarrow\left|5x\right|\le2\)( x \(\ge0\))
Mà không có số x nào nhân với 5 bé hơn hoặc bằng 2
\(\Rightarrow\)x không có giá trị thỏa mãn
c đề bài sai, chỉ tìm x chứ làm gì có y
d, \(\left(x-3\right).\left(2y+1\right)=7\)
TH1:
x - 3 = 1
x = 1 + 3
x = 4
2y + 1 = 7
2y = 7 - 1 = 6
y = 6 : 2 = 3
TH2:
x - 3 = 7
x = 7 + 3 = 10
2y + 1 = 1
2y = 1 - 1 = 0
y = 0 : 2 = 0
TH3:
x - 3 = -1
x = -1 + 3
x = 2
2y+ 1 = -7
2y = -7 - 1 = -8
y = (-8) : 2 = -4
TH4:
x - 3 = -7
x = -7 + 3
x = -4
2y + 1 = -1
2y = (-1) - 1
2y = -2
y = (-2) : 2 = -1
Vậy ......
thì ra mày đi làm những bài này ? gg , theo t , t dạy m cách giải pt bậc 2 3 4 " siêu tốc " :)
Tìm x , y thỏa mãn :
a) \(\frac{1}{2}\times(\frac{3}{4}x-\frac{1}{2})^{2018}+\frac{2017}{2018}\times/\frac{4}{5}y+\frac{6}{25}/\le0\)0
b) \(2017\times/2x-y/+2018\times(y-4)^{2017}\le0\)
Bài 1: Tìm x, y, z biết
\(\left|x-\frac{1}{2}\right|+\left|2y-\frac{1}{3}\right|+\left|4z+5\right|\le0\)
Bài 2: Viết các biểu thức sau dưới dạng thu gọn
A = |x - 1| + x + 3
B = 2x - |2x + 3|
B1:
Vì \(\hept{\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|2y-\frac{1}{3}\right|\ge0\\\left|4z+5\right|\ge0\end{cases}\left(\forall x,y,z\right)}\Rightarrow\left|x-\frac{1}{2}\right|+\left|2y-\frac{1}{3}\right|+\left|4z+5\right|\ge0\left(\forall x,y,z\right)\)
Mà theo đề bài, \(\left|x-\frac{1}{2}\right|+\left|2y-\frac{1}{3}\right|+\left|4z+5\right|\le0\) nên dấu "=" xảy ra khi:
\(\left|x-\frac{1}{2}\right|=\left|2y-\frac{1}{3}\right|=\left|4z+5\right|=0\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{6}\\z=-\frac{5}{4}\end{cases}}\)
B2:
a) Nếu \(x< 1\) => \(A=1-x+x+3=4\)
Nếu \(x\ge1\) => \(A=x-1+x+3=2x+2\)
b) Nếu \(x< -\frac{3}{2}\) => \(B=2x+2x+3=4x+3\)
Nếu \(x\ge-\frac{3}{2}\) => \(B=2x-2x-3=-3\)
Bài 1.
Ta có \(\hept{\begin{cases}\left|x-\frac{1}{2}\right|\ge0\forall x\\\left|2y-\frac{1}{3}\right|\ge0\forall y\\\left|4z+5\right|\ge0\forall z\end{cases}}\Rightarrow\left|x-\frac{1}{2}\right|+\left|2y-\frac{1}{3}\right|+\left|4z+5\right|\ge0\forall x,y,z\)
Kết hợp với đề bài => Chỉ xảy ra trường hợp \(\left|x-\frac{1}{2}\right|+\left|2y-\frac{1}{3}\right|+\left|4z+5\right|=0\)
\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\2y-\frac{1}{3}=0\\4z+5=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{6}\\z=-\frac{5}{4}\end{cases}}\)
Bài 2.
A = | x - 1 | + x + 3
Với x < 1 => A = -( x - 1 ) + x + 3 = -x + 1 + x + 3 = 4
Với x ≥ 1 => A = ( x - 1 ) + x + 3 = x - 1 + x + 3 = 2x + 2
B = 2x - | 2x + 3 |
Với x < -3/2 => B = 2x - -( 2x + 3 ) = 2x + ( 2x + 3 ) = 2x + 2x + 3 = 4x + 3
Với x ≥ -3/2 => B = 2x + -( 2x + 3 ) = 2x - ( 2x + 3 ) = 2x - 2x - 3 = -3
tìm x,y biết:
a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
b) \(\left(\dfrac{1}{2}.x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\)( do \(x^2\ge0,\left(y-\dfrac{1}{10}\right)^4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
b) \(\left(\dfrac{1}{2}.x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\)( do \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0,\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
\(a,\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\\ b,\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\end{matrix}\right.\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
Mà \(x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x;y\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=0\\\left(y-\dfrac{1}{10}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(0;\dfrac{1}{10}\right)\)
b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\forall x;y\)
\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}=0\\\left(y^2-\dfrac{1}{4}\right)^{10}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(10;\dfrac{1}{2}\right);\left(10;-\dfrac{1}{2}\right)\right\}\)
b\(|x+\frac{9}{2}|+|y+\frac{4}{3}|+|z+\frac{7}{2}|\le0\)
Tìm x, y, z
Ta có: \(\left|x+\frac{9}{2}\right|\ge0\)
\(\left|y+\frac{4}{3}\right|\ge0\)
\(\left|z+\frac{7}{2}\right|\ge0\)
Mà \(\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\le0\)
\(\Rightarrow\hept{\begin{cases}x+\frac{9}{2}=0\\y+\frac{4}{3}=0\\z+\frac{7}{2}=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x=\frac{-9}{2}\\y=\frac{-4}{3}\\z=-\frac{7}{2}\end{cases}}\)
Vậy....................
Ta có:\(\hept{\begin{cases}\left|x+\frac{9}{2}\right|\ge0\forall x\\\left|y+\frac{4}{3}\right|\ge0\forall y\\\left|z+\frac{7}{2}\right|\ge0\forall z\end{cases}}\)
\(\Rightarrow\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\ge0\forall x,y,z\)
Dấu "="xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{9}{2}\right|=0\\\left|y+\frac{4}{3}\right|=0\\\left|z+\frac{7}{2}\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+\frac{9}{2}=0\\y+\frac{4}{3}=0\\z+\frac{7}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-9}{2}\\y=\frac{-4}{3}\\z=\frac{-7}{2}\end{cases}}\)
Vậy.... thỏa mãn đề bài
BÀI 1: tìm x biết : \(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)
BÀI 2: tìm số tự nhiên x thỏa mãn: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{16}{34}\)
BÀI 3: Cho x;y thỏa mãn : \(\left(x-2014\right)^{2010}+\left(y-2010\right)^{2014}\le0\)
bài 1
[(x+2)/1010]+ [(x+2)/1111]= [(x+2)/1212]+[(x+2)/1313]
=>[(x+2)/1010]+[(x+2)/1111] - [(x+2)/1212]-[(x+2)/1313] = 0
=>(x+2).[(1/1010)+(1/1111)-(1/1212)-(1/1313)=0
Vì [(1/1010)+(1/1111)-(1/1212)-(1/1313)] khác 0
=>x+2=0
=>x=-2
Bài 1 : -2
Bài 2 : 15
Bải 3 : x =2014 ; y = 2010