\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)

\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)

\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)

=> x - 29 = 0

=> x = 29.

\(\frac{1}{x+y+z}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{3}\)

rồi quy đồng biến            đổi ra : (x + y ) ( y + z ) ( z + x ) = 0

=> x = -y hoặc y = -z hoặc z = -x

kết hợp với 2x² + y = 1 rồi giải ra sau đó thay vào x + y + z = 3 rồi tìm ra x,y,z

như thế bạn làm nhá

\(M=\frac{x-2-\sqrt{x}-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}}\)

a.Ta co:\(x^2-x=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=1\left(n\right)\end{cases}}\)

\(\Rightarrow M=\frac{1-2}{1}=-1\)

b.De \(M\in Z\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}}\in Z\Rightarrow\sqrt{x}-2⋮\sqrt{x}\Rightarrow x=4\)

Mình cảm ơn bạn nhiều nha ^^

By Titu's Lemma we easy have:

\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)

\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

1) có \(2y\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)

\(\Rightarrow\left(\sqrt{xy}+\frac{1}{4\sqrt{xy}}\right)^2+\frac{15}{16xy}+\frac{1}{2}\ge\frac{15}{16}\cdot4+\frac{1}{2}=\frac{17}{4}\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

\(\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+zx+1}\)

\(=\frac{xyz}{x\left(1+y+yz\right)}+\frac{1}{1+y+yz}+\frac{xyz}{xz\left(1+y+yz\right)}\)

\(=\frac{yz}{1+y+yz}+\frac{1}{1+y+yz}+\frac{y}{1+y+yz}\)

\(=\frac{1+y+yz}{1+y+yz}\)

\(=1\)

tích đi rồi t làm 

9 T I C H  sai buồn

\(A=\frac{\sqrt{x^3}}{\sqrt{xy}-2y}-\frac{2x}{x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}}.\frac{1-x}{1-\sqrt{x}}..\)

nhờ vào năng lực rinegan tối hậu của ta , ta có thể dễ dàng nhìn thấy mẫu chung 

\(x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}=\sqrt{x}\left(\sqrt{x}-2\sqrt{xy}\right)+\left(\sqrt{x}-2\sqrt{y}\right)=\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+1\right)\)

\(A=\frac{\sqrt{x^3}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}-\frac{2x\left(x-1\right)}{\left(\sqrt{x}-2\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}.\)

\(\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)

\(A=\frac{\sqrt{x^3}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\sqrt{x}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\left(\sqrt{x}-2\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x}{\sqrt{y}}\)

b) thay y=625 vào ta được

\(\frac{x}{\sqrt{625}}=\frac{x}{25}< 0.2\Leftrightarrow x< 5\)

vậy   \(0< x< 5\)

Ta có : 

\(\frac{-16}{32}=\frac{-16:16}{32:16}=\frac{-1}{2}\)

+)\(\frac{-1}{2}=\frac{x}{-10}\)

=> (-10) x (-1) = X x 2

=> 10 = X x 2

=> X = 10 : 2 

=> X = 5

+) \(\frac{-1}{2}=\frac{-7}{y}\)

=> (-1) x Y = (-7) x 2

=> -Y = -14

=> Y = 14

+)\(\frac{-1}{2}=\frac{z}{24}\)

=> (-1) x 24 = Z x 2

=> -24 = Z x 2

=> Z = -24 : 2

=> Z = -12

Kết luận : X = 5

                Y = 14

                Z = 12

a) \(P=\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}=\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\)

b)

\(\frac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{y}}\\ =\frac{\left(1+\sqrt{x}\right)+\sqrt{y}\left(1+\sqrt{x}\right)}{1+\sqrt{y}}\\ =\frac{\left(1+\sqrt{y}\right)\left(1+\sqrt{x}\right)}{1+\sqrt{y}}\\ =1+\sqrt{x}\)