1) ĐK:x\(\ge\frac{1}{2}\)

PT\(\Leftrightarrow\sqrt{2x-1}=x\)

\(\Leftrightarrow\begin{cases}x\ge0\\2x-1=x^2\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge0\\x=1\end{cases}\)

\(\Leftrightarrow x=1\)   (thỏa mãn)

\(A=\frac{\left(3+\sqrt{5}\right)^2+\left(3-\sqrt{5}\right)^2}{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(A=\frac{18+10}{4}\)

\(A=7\)

\(B=\sqrt{9-3\times2\sqrt{3}+3}+\sqrt{12-2\times3\times2\sqrt{3}+9}\)

\(B=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^3}\)

\(B=\left|3-\sqrt{3}\right|+\left|2\sqrt{3}-3\right|\)

\(B=3-\sqrt{3}+2\sqrt{3}-3\)

\(B=\sqrt{3}\)

a) ĐKXĐ: x\(\ge\)-3

PT\(\Leftrightarrow\sqrt{\left(x+7\right)\left(x+3\right)}=3\sqrt{x+3}+2\sqrt{x+7}-6\)

Đặt \(\left(\sqrt{x+3},\sqrt{x+7}\right)=\left(a,b\right)\)                 \(\left(a,b\ge0\right)\)

PT\(\Leftrightarrow ab=3a+2b-6\Leftrightarrow a\left(b-3\right)-2\left(b-3\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(b-3\right)=0\Leftrightarrow\orbr{\begin{cases}a=2\\b=3\end{cases}}\)(TM ĐK)

TH 1: a=2\(\Leftrightarrow\sqrt{x+3}=2\Leftrightarrow x+3=4\Leftrightarrow x=1\)(tm)

TH 2: b=3\(\Leftrightarrow\sqrt{x+7}=3\Leftrightarrow x+7=9\Leftrightarrow x=2\)(tm)

Vậy tập nghiệm phương trình S={1; 2}

Vũ Minh Tuấn, Băng Băng 2k6, Hoàng Tử Hà, đề bài khó wá, Lê Gia Bảo, Aki Tsuki, Nguyễn Việt Lâm,

Lê Thị Thục Hiền, Nguyễn Trúc Giang, Học 24h, @tth_new, @Akai Haruma

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thanks!

1)

a) \(\left\{{}\begin{matrix}2x-y=5\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-y+x+y=5+4\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x=9\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Vậy (x;y)=(3;1)

b) \(16x^5-8x^3+x=0\Leftrightarrow x\left(16x^4-8x^2+1\right)=0\Leftrightarrow x\left[\left(4x^2\right)^2-2.4x^2.1+1^2\right]=0\Leftrightarrow x\left(4x^2-1\right)^2=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\4x^2-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=\frac{\pm1}{2}\end{matrix}\right.\)

Vậy S={\(-\frac{1}{2};0;\frac{1}{2}\)}

2)

A=\(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{4}+\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{5}+1}{5-1}=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}-1+\sqrt{5}+1}{4}=\frac{2\sqrt{5}}{4}=\frac{\sqrt{5}}{2}\)

B=\(\frac{4}{3+\sqrt{5}}-\frac{8}{1+\sqrt{5}}+\frac{15}{\sqrt{5}}=\frac{4\left(3-\sqrt{5}\right)}{9-5}-\frac{8\left(1-\sqrt{5}\right)}{1-5}+3\sqrt{5}=\frac{4\left(3-\sqrt{5}\right)}{4}-\frac{8\left(\sqrt{5}-1\right)}{4}+3\sqrt{5}=3-\sqrt{5}-2\sqrt{5}+2+3\sqrt{5}=5\)

1.

ĐK: \(-1\le x\le4\)

Đặt \(\sqrt{x+1}+\sqrt{4-x}=t\left(t\ge0\right)\)

\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{t^2-5}{2}\)

\(PT\Leftrightarrow t+\frac{t^2-5}{2}=5\Rightarrow t^2+2t-15=0\) \(\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)

\(t=3\Rightarrow\sqrt{-x^2+3x+4}=2\) \(\Leftrightarrow-x^2+3x+4=4\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (tm)

2.

ĐK:\(x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t\ge0\right)\)

\(\Rightarrow2\sqrt{x^2-16}=t^2-2x\)

\(PT\Leftrightarrow t=2x-12+t^2-2x\)

\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\) Giải tiếp như trên.