\(P=2\sqrt{x-3}+7\sqrt{4x-12}-3\sqrt{25x-75}+8\sqrt{\frac{x-3}{4}}+3\left(x\ge3\right)\)

\(P=2\sqrt{x-3}+14\sqrt{x-3}-15\sqrt{x-3}+4\sqrt{x-3}+3\)

\(P=5\sqrt{x-3}+3\ge3\)

Dấu = xảy ra

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Vậy ...

a=2³⁰+2²⁰²⁰+4ⁿ=4¹⁵+4¹⁰¹⁰+4ⁿ=4¹⁵(1+4⁹⁹⁵+4^n-15) mà 4¹⁵ là số cp suy ra (1+4⁹⁹⁵+4^n-15)là số cp

ta có: 1+4⁹⁹⁵+4^n-15=2^2n-30+2.2¹⁹⁸⁹+1=(2^2n-30+1)²

đề 1+4⁹⁹⁵+4^n-15=(2^n-15)²+2.2¹⁹⁸⁹+1=(2^n-15+1)² là số cp thì n-15=1989 suy ra n=1974

nếu sai thì sorry bạn nha

\(P=\frac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{1-\sqrt{x}}\)(ĐK: \(x\ge0,x\ne1\)) 

\(=\frac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\left(3\sqrt{x}+8\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{3\sqrt{x}+8}{\sqrt{x}+2}\)

\(P< \frac{15}{4}\)

\(\Leftrightarrow\frac{3\sqrt{x}+8}{\sqrt{x}+2}< \frac{15}{4}\)

\(\Leftrightarrow4\left(3\sqrt{x}+8\right)< 15\left(\sqrt{x}+2\right)\)

\(\Leftrightarrow3\sqrt{x}>2\)

\(\Leftrightarrow x>\frac{4}{9}\)

Vậy \(x>\frac{4}{9},x\ne1\)thì \(P< \frac{15}{4}\).

\(P=\frac{3\sqrt{x}+8}{\sqrt{x}+2}=3+\frac{2}{\sqrt{x}+2}\le3+\frac{2}{0+2}=4\)

Dấu \(=\)khi \(x=0\).

em

chịu 

nha 

anh

\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}\)

\(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}\)

\(a+\sqrt{a}+1\)

Bổ sung: ĐKXĐ: \(a\ge0;a\ne1\)

ĐK: \(x\ge-1\).

\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)

\(\Leftrightarrow\sqrt[3]{x-2}-1+\sqrt{x+1}-2=0\)

\(\Leftrightarrow\frac{x-2-1}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{x+1-4}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt[3]{\left(x-2\right)^2+\sqrt[3]{x-2}+1}}+\frac{1}{\sqrt{x+1}+2}\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)(thỏa mãn)

https://h7.net/hoi-dap/toan-10/giai-phuong-trinh-can-bac-3-x-2-can-x-1-3-faq399261.html

link tham khảo nhé

chúc bạn hok tốt

Giải phương trình sau: ∛(x-2) + √(x+1)=3

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

\(A=\left(\frac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\left(\frac{\sqrt{x}-3}{\sqrt{x}}\right)\)

\(A=\frac{2\sqrt{x}}{\sqrt{x}+3}.\frac{1}{\sqrt{x}}\)

\(A=\frac{2}{\sqrt{x}+3}.1=\frac{2}{\sqrt{x}+3}\)

\(B=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}+\frac{6-7\sqrt{x}}{x-4}\right)\left(\sqrt{x}-2\right)\) \(ĐKXĐ:x\ne4;x\ge0\)

\(B=\left(\frac{\sqrt{x}.\left(\sqrt{x}+2\right)+1.\left(\sqrt{x}-2\right)+6-7\sqrt{x}}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}\right).\left(\sqrt{x}+2\right)\)

\(B=\frac{x+2\sqrt{x}+\sqrt{x}-2+6-7\sqrt{x}}{\sqrt{x}-2}\)

\(B=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}\)

\(B=\frac{\left(\sqrt{x}\right)^2-2.2.\sqrt{x}+2^2}{\sqrt{x}-2}\)

\(B=\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-2}\)

\(B=\sqrt{x}-2\)

Vậy...

mình ghi sai đề \(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)

\(\frac{\left(3\sqrt{2}-\sqrt{10}\right)\sqrt{6+2\sqrt{5}}+\left(3\sqrt{2}+\sqrt{10}\right)\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)

\(\frac{\left(3\sqrt{2}-\sqrt{10}\right)\left(\sqrt{5}+1\right)+\left(3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{5}-1\right)}{\sqrt{2}}\)

\(\frac{3\sqrt{10}-5\sqrt{2}+3\sqrt{2}-\sqrt{10}+3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}}{\sqrt{2}}\)

\(\frac{4\sqrt{10}}{\sqrt{2}}=4\sqrt{5}\)