=> 3( xy- 2 ) = 4 y => 3xy -4y = 6

=> \(y=\frac{6}{3x-4}\)

=> 3x -4 thuộc U(6) = { -6;-3;-2;-1;1;2;3;6}

Nguyen Huu The - ko giải được thì thôi , lúc nào cug nói " sorry , mới lớp 6 thui "

Nguyen Huu The lớp 7 đó, nhưng trình độ lớp 6 thui

câu 1 thiếu đề

câu 2:

Ta có: 2¹⁵⁰=(2⁶)²⁵=64²⁵

          3¹⁰⁰=(3⁴)²⁵=81²⁵

Vì 64²⁵<81²⁵ nên 2¹⁵⁰<3¹⁰⁰

x o dau vay???

2^150 =(2^3)^50=8^ 50

3^100= (3^2)^50 =9^50

ma 8^50< 9^50=> 2^150<3^100

Ta có: \(\frac{1}{1+x}\ge\left(1-\frac{1}{1+y}\right)+\left(1-\frac{1}{1+z}\right)\ge2\sqrt{\frac{yz}{\left(1+y\right)\left(1+z\right)}}\)

Tương tự cho 2 cái còn lại:

\(\frac{1}{1+y}\ge2\sqrt{\frac{xz}{\left(z+1\right)\left(x+1\right)}};\frac{1}{1+z}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)

Nhân theo vế ta được:

\(\frac{1}{1+x}\cdot\frac{1}{1+y}\cdot\frac{1}{1+z}\ge\frac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)\(\Rightarrow xyz\le\frac{1}{8}\)

Dấu = khi \(\hept{\begin{cases}x=y=z\\\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=2\end{cases}}\Leftrightarrow x=y=z=\frac{1}{2}\)

em mới lớp 1 àk

Huỳnh Đăng Trình điu vừa thui

lam an de sao hay vay

Bài 2: 

a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)

b: Thay x=1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)

Thay x=-1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)

c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)

=>6(x-2)=-1/2

=>x-2=-1/12

hay x=23/12

Ta có \(\frac{1}{P}=\frac{\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)^2}{x^3y^3}=\frac{x+yz}{y}\cdot\frac{y+zx}{x}\cdot\frac{\left(z+xy\right)^2}{x^2y^2}\)

\(=\left(\frac{x}{y}+z\right)\left(\frac{y}{x}+z\right)\left(\frac{z}{xy}+1\right)^2=\left[1+\left(\frac{x}{y}+\frac{x}{y}\right)z+x^2\right]\left(\frac{z}{xy}+1\right)^2\ge\left(1+2x+x^2\right)\)\(\left[\frac{4x}{\left(x+y\right)^2}+1\right]^2\)\(=\left(z+1\right)^2\left[\frac{4z}{\left(z-1\right)^2}+1\right]^2=\left[\frac{4z\left(z+1\right)}{\left(z-1\right)^2}+1\right]^2=\left[6+\frac{12}{z-1}+\frac{8}{\left(z-1\right)^2}+z-1\right]^2\)

\(=\left[6+\frac{12}{z-1}+\frac{3\left(z-1\right)}{4}+\frac{8}{\left(z-1\right)^2}+\frac{z-1}{8}+\frac{z-1}{8}\right]\)

Áp dụng BĐT Cosi ta có:

\(\frac{1}{P}\ge\left[6+2\sqrt{\frac{12}{z-1}\cdot\frac{3\left(z-1\right)}{3}}+3\sqrt[3]{\frac{8}{\left(z-1\right)^2}\cdot\frac{z-1}{8}\cdot\frac{z-1}{8}}\right]^2=\frac{729}{4}\)

\(\Rightarrow P\le\frac{4}{729}\). dấu "=" xảy ra <=> \(\hept{\begin{cases}x=y=2\\z=5\end{cases}}\)

Câu 1) a) ĐKXĐ \(x\ge0,\)\(x\ne4\)A=\(\frac{x+2\sqrt{x}-4}{2\left(x-4\right)}\)b) Mình chưa làm được       Câu 2) a) ĐKXĐ \(x>0,\)\(x\ne4\)A=\(\frac{\sqrt{x}-1}{\sqrt{x}}\)b) Để a<\(\frac{1}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}< \frac{1}{2}\)\(\Rightarrow x< 1\)\(\Rightarrow0< x< 1\)thỏa mãn bài toán    c) Ta có A=\(\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\), để A \(\in Z\)\(\Rightarrow\sqrt{x}\inƯ\left(1\right)\), \(\Rightarrow x=1\)( thỏa mãn ĐK)

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