1. \(\left(2x+5y+1\right)\left(2^{\left|x\right|}+y+x^2+x\right)=105\)

\(\Rightarrow\) \(2x+5y+1\) và \(2^{\left|x\right|}+y+x^2+x\) cùng lẻ

Từ \(2x+5y+1\) lẻ => y chẵn

Mà \(x^2+x=x\left(x+1\right)\) chẵn \(\forall x\in Z\) nên \(y+x^2+x\) chẵn

Mặt khác \(2^{\left|x\right|}+y+x^2+x\) lẻ => \(2^{\left|x\right|}\) lẻ

=> x = 0, y tự tìm.

4. Đặt \(4P^2+1=k^2\) \(\left(k\in Z\right)\)

\(\Leftrightarrow k^2-4P^2=1\)

\(\Leftrightarrow\left(k-2P\right)\left(k+2P\right)=1\)

Xét các ước của 1 để tìm k và P (lưu ý P là số nguyên tố)

Ta có \(x+y+z\ge3\sqrt[3]{xyz}=3.1=3\)

Theo BĐT Cauchy-Schwarz dạng Engel:

\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)+3}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)+\left(x+y+z\right)}=\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3}{2}\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{y+1}=\dfrac{y}{z+1}=\dfrac{z}{x+1}\\xyz=1\end{matrix}\right.\)

\(\Leftrightarrow x=y=z=1\)

3/ Ta viết lại pt thứ nhất của hệ

\(x^2-x\left(2y-3\right)+y^2-3y-4=0\)

\(\Leftrightarrow x^2-x\left(2y-3\right)+\dfrac{4y^2-12y+9}{4}-\dfrac{25}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{2y+3}{2}\right)^2-\left(\dfrac{5}{2}\right)^2=0\)

\(\Leftrightarrow\left(x-y-4\right)\left(x-y+1\right)=0\)

Bạn làm tiếp được chứ?

4/ Viết lại pt thứ 2 của hệ

\(\left(y+\sqrt{x}\right)^2-\left(y\sqrt{x}\right)^2=0\)

\(\Leftrightarrow\left(y-\sqrt{x}-y\sqrt{x}\right)\left(y-\sqrt{x}+y\sqrt{x}\right)=0\)

BCNN(m;n)=m

VD:m=9 ;n=3

VÌ 9 chia hết cho 3=>BCNN(9;3)=9

1/ \(\left\{{}\begin{matrix}x^3+y^3=1\left(1\right)\\x^2y+2xy^2+y^3=2\left(2\right)\end{matrix}\right.\)

Lấy (1). 2 - (2) ta được:

\(2x^3+y^3-x^2y-2xy^2=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left(2x-y\right)=0\)

Đến đây dễ rồi nhé ^^

2/ Ta viết lại pt thứ 2 của hệ:

\(y^2-4\left(x+2\right)y+16+16x-5x^2=0\)

\(\Leftrightarrow y^2-4\left(x+2\right)y+4\left(x+2\right)^2-9x^2=0\)

\(\Leftrightarrow\left[y-2\left(x+2\right)\right]^2-\left(3x\right)^2=0\)

\(\Leftrightarrow\left(x+y-4\right)\left(y-5x-4\right)=0\)

Bạn làm tiếp nhé!

ĐK: \(x\ge1\)

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

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

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

\(\ge\left|1-\sqrt{x-1}+\sqrt{x-1}+1\right|=2\)

Đẳng thức xảy ra \(\Leftrightarrow\left(1-\sqrt{x-1}\right)\left(\sqrt{x-1}+1\right)\ge0\)

\(\Leftrightarrow1\le x\le2\)

\(S=\dfrac{\sqrt[3]{\left(a-2\right)\left(b-3\right)}}{a+b}\)

\(\Rightarrow S.\sqrt[3]{5}=\dfrac{\sqrt[3]{\left(a-2\right)\left(b-3\right).5}}{a+b}\)

\(\le\dfrac{\dfrac{\left(a-2\right)+\left(b-3\right)+5}{3}}{a+b}=\dfrac{\dfrac{a+b}{3}}{a+b}=\dfrac{1}{3}\)

\(\Rightarrow S\le\dfrac{1}{3}:\sqrt[3]{5}=\dfrac{1}{3\sqrt[3]{5}}\)

Đẳng thức xảy ra \(\Leftrightarrow a-2=b-3=5\Leftrightarrow\left\{{}\begin{matrix}a=7\\b=8\end{matrix}\right.\)

\(P=\dfrac{\sqrt{a-2}}{a}+\dfrac{\sqrt[3]{b-3}}{b}+\dfrac{\sqrt[4]{c-6}}{c}\)

\(=\dfrac{\sqrt{\left(a-2\right).2}}{a\sqrt{2}}+\dfrac{\sqrt[3]{\left(b-3\right).\dfrac{3}{2}.\dfrac{3}{2}}}{b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{\sqrt[4]{\left(c-6\right).2.2.2}}{c\sqrt[3]{8}}\)

\(\le\dfrac{a-2+2}{2a\sqrt{2}}+\dfrac{b-3+\dfrac{3}{2}+\dfrac{3}{2}}{3b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{c-6+2+2+2}{4c\sqrt[4]{8}}\)

\(=\dfrac{a}{2a\sqrt{2}}+\dfrac{b}{3b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{c}{4c\sqrt[4]{8}}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{\dfrac{9}{4}}}+\dfrac{1}{4\sqrt[4]{8}}\)

Vậy \(P_{max}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{\dfrac{9}{4}}}+\dfrac{1}{4\sqrt[4]{8}}\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-2=2\\b-3=\dfrac{3}{2}\\c-6=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=\dfrac{9}{2}\\c=8\end{matrix}\right.\)