1C 2C 3D 4D 5A 6B 7B 8C 9A 10D 11D 12A 13B

Đề thế này phải ko bạn: 

Chứng minh rằng: \(x^5+y^5\ge x^4.y+x.y^4\)với \(x,y\ne0\)và\(x+y\ge0\)

bạn vào fx viết lại đề đi nha, sai đề rùi

Ta có: \(x^5+y^5\ge x^4.y+x.y^4\)(1)

<=>\(x^5+y^5-x^4.y-x.y^4\ge0\)

<=>\(\left(x^5-x^4.y\right)-\left(x.y^4-y^5\right)\ge0\)

<=>\(x^4.\left(x-y\right)-y^4.\left(x-y\right)\ge0\)

<=>\(\left(x^4-y^4\right).\left(x-y\right)\ge0\)

<=>\(\left[\left(x^2\right)^2-\left(y^2\right)^2\right].\left(x-y\right)\ge0\)

<=>\(\left(x^2+y^2\right).\left(x^2-y^2\right).\left(x-y\right)\ge0\)

<=>\(\left(x^2+y^2\right).\left(x+y\right).\left(x-y\right).\left(x-y\right)\ge0\)

<=>\(\left(x^2+y^2\right).\left(x+y\right).\left(x-y\right)^2\ge0\)

Vì \(x^2+y^2\ge0,\left(x-y\right)^2\ge0\)

=>(1)<=>\(x+y\ge0\)(2)

Vì \(x+y\ge0\)(theo giả thiết)

=>(2) đúng với mọi x,y

Vì các dấu"<=>" có giá trị như nhau

=>(1) đúng với mọi x,y

=>ĐPCM

|x-2|=x

x-2=x

x+x=2

2x=2

\(\Rightarrow x=1\)

35 x 12 + 65 x 13 = 35 x 12 + 65 x 12 + 65 = 12 x ( 35 + 65 ) + 65 = 12 x 100 + 65 = 1200 + 65 = 1265

35x12+65x13=420+845=1265

tk cho mk nha

35 x 12 + 65 x 13=1265

=>-13/9  < X < -11/18

=>-26/18<X<-11/18

=> X E {-25/18;-24/18;......;-11/18}

 K NHA

ỦNG HỘ MK NHA

Đề bài là tìm số hữu tỉ x nha

\(\frac{3x-1}{8}+\frac{3x+18}{11}=\frac{3x}{7}+\frac{3x+20}{13}\)

\(\Rightarrow\frac{1001\left(3x-1\right)}{8008}+\frac{728\left(3x+18\right)}{8008}=\frac{1144.3x}{8008}+\frac{616\left(3x+20\right)}{8008}\)

\(\Rightarrow3003x-1001+2184x+13104x=3432x+1848x+12320\)\

\(\Rightarrow\)\(19111x=13321\Rightarrow x=\frac{13321}{19111}\)

\(\frac{3\times-1}{8}+\frac{3\times+18}{11}=\frac{3\times}{7}+\frac{3\times+20}{13}\)

\(\Rightarrow\frac{3\times-1}{8}-\frac{3\times}{7}=\frac{3\times+20}{13}-\frac{3\times+18}{11}\)

\(\Rightarrow\frac{3\times-1}{8}+1-\frac{3\times}{7}-1=\frac{3\times+20}{13}-1-\frac{3\times+18}{11}+1\)

\(\Rightarrow\left(\frac{3\times-1}{8}+1\right)-\left(\frac{3\times}{7}+1\right)=\left(\frac{3\times+20}{13}-1\right)-\left(\frac{3\times+18}{11}-1\right)\)

\(\Rightarrow\frac{3\times+7}{8}-\frac{3\times+7}{7}=\frac{3\times+7}{13}-\frac{3\times+7}{11}\)

\(\Rightarrow\left(3\times+7\right)\cdot\left(\frac{1}{8}-\frac{1}{7}\right)=\left(3\times+7\right)\cdot\left(\frac{1}{13}-\frac{1}{11}\right)\)

\(\Rightarrow\left(3\times+7\right)\cdot\left(\frac{-1}{56}\right)=\left(3\times+7\right)\cdot\left(\frac{-2}{143}\right)\)

\(\Leftrightarrow3\times+7=0\)

\(\Leftrightarrow3\times=-7\)

\(\Leftrightarrow\times=\frac{-7}{3}\)

\(\frac{-7}{3}\inℚ\)

Vậy \(\times=\frac{-7}{3}\)

(3^3)^2 + 2^x = 5 x ( 45 +2^2 x 25 )+ 8 

=> 9^2 + 2^x = 5 x( 45 + 4 x 25) + 8 

=> 81 + 2^x = 5 x ( 45 + 100 0 + 8 

=> 81 + 2^x = 5 x 145 + 8 

=> 81 + 2^x = 725 + 8 

=> 81 + 2^x = 733

=> 2^x = 733- 81 

=>2^x =652

=> không có giá trị x thỏa mãn

sau khi rút gọn
2^x = 225+500+8-729 = 4
=>  2^x=4  
=>  x = 2