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 \(x=0,7\)\(y=-6,3\)

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 " :)

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

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 \(\left(x;y\right)\in\left\{\left(10;\dfrac{1}{2}\right);\left(10;-\dfrac{1}{2}\right)\right\}\)

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

[(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: x=-2

Bài 2:x=17

Bài 3:x=2014

y=2010

Bài 1 : -2

Bài 2 : 15

Bải 3 : x =2014 ; y = 2010