a)

\(\sqrt{x}=4\Rightarrow x=4^2=16\)

c) \(x\in\varnothing\)

e)  \(\sqrt{x}=6,25\Rightarrow x=\left(6,25\right)^2=39,0625\)

b) \(\sqrt{x}=\sqrt{7}\Rightarrow x=7\)

d) \(\sqrt{x}=0\Rightarrow x=0\)

Cách đánh đề độc lạ ghê:v

a: =>x=16

b: =>x=7

c: =>x thuộc rỗng

d: =>x=0

e: =>x=(25/4)^2=625/16

\(\left(a\right):2x-7\sqrt{x}+3=0\left(x\ge0\right)\\ < =>\left(2x-6\sqrt{x}\right)-\left(\sqrt{x}-3\right)=0\\ < =>2\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)=0\\ < =>\left(2\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\\ =>\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{4}\left(TM\right)\\x=9\left(TM\right)\end{matrix}\right.\)

\(\left(b\right):3\sqrt{x}+5< 6\\ < =>3\sqrt{x}< 1\\ < =>\sqrt{x}< \dfrac{1}{3}\\ < =>0\le x< \dfrac{1}{9}\)

\(\left(c\right):x-3\sqrt{x}-10< 0\\ < =>\left(x-5\sqrt{x}\right)+\left(2\sqrt{x}-10\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}-5\right)+2\left(\sqrt{x}-5\right)< 0\\ < =>\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}+2>0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}0\le x< 25\\x\ge0\end{matrix}\right.< =>0\le x< 25\)

\(\left(d\right):x-5\sqrt{x}+6=0\left(x\ge0\right)\\ < =>\left(x-2\sqrt{x}\right)-\left(3\sqrt{x}-6\right)=0\\ < =>\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=0\\ < =>\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\left(TM\right)\)

\(\left(e\right):x+5\sqrt{x}-14< 0\\ < =>\left(x+7\sqrt{x}\right)-\left(2\sqrt{x}+14\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}+7\right)-2\left(\sqrt{x}+7\right)< 0\\ < =>\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}+7>0\\\sqrt{x}-2< 0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.< =>0\le x< 4\)

Bài 8:

\(M=1+\frac{4}{\sqrt{x}+1}\)

Để $M$ nguyên thì $\frac{4}{\sqrt{x}+1}$ nguyên 

Đặt $\frac{4}{\sqrt{x}+1}=t$ với $t$ là số nguyên dương 

$\Rightarrow \sqrt{x}+1=\frac{4}{t}$

$\sqrt{x}=\frac{4}{t}-1=\frac{4-t}{t}\geq 0$

$\Rightarrow 4-t\geq 0\Rightarrow t\leq 4$

Mà $t$ nguyên dương suy ra $t=1;2;3;4$

Kéo theo $x=9; 1; \frac{1}{9}; 0$

Kết hợp đkxđ nên $x=0; \frac{1}{9};9$

Bài 9:

$P=1+\frac{5}{\sqrt{x}+2}$

Để $P$ nguyên thì $\frac{5}{\sqrt{x}+2}$ nguyên 

Đặt $\frac{5}{\sqrt{x}+2}=t$ với $t\in\mathbb{Z}^+$

$\Leftrightarrow \sqrt{x}+2=\frac{5}{t}$

$\Leftrightarrow \sqrt{x}=\frac{5-2t}{t}\geq 0$

Với $t>0\Rightarrow 5-2t\geq 0$

$\Leftrightarrow t\leq \frac{5}{2}$

Vì $t$ nguyên dương suy ra $t=1;2$

$\Rightarrow x=9; \frac{1}{4}$ (thỏa đkxđ)

Bài 8: 

Để M nguyên thì \(\sqrt{x}+5⋮\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(4\right)\)

\(\Leftrightarrow\sqrt{x}+1\in\left\{1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;1;3\right\}\)

hay \(x\in\left\{0;1;9\right\}\)

\(7-\sqrt{x}=0\)

\(\sqrt{x}=7\)

\(\Rightarrow x=7^2\)

\(\Rightarrow x=49\)

vậy \(x=49\)

P/S: fan khởi my

\(7-\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}=7-0=7\)

\(\Leftrightarrow x=7\cdot7\)

\(\Leftrightarrow x=49\)

Ta có:

\(7-\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}=7\)

\(\Rightarrow x=49\)

P/s tham khảo nha

\(x-9\sqrt{x}+14=0\Leftrightarrow x-2\sqrt{x}-7\sqrt{x}+14=0\)

                                        \(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)-7\left(\sqrt{x}-2\right)=0\)

                                         \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-7\right)=0\)

                                           \(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=49\end{cases}}}\)

Vậy x = 4 hoặc x = 49

\(\sqrt{x^2-10x+25}=7-2x\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=7-2x\)

\(\Leftrightarrow\left|x-5\right|=7-2x\)(1)

Nếu \(x-5\ge0\Rightarrow x\ge5\) thì (1) trở thành: x-5=7-2x <=> 3x=12 <=> x=4 (loại)

Nếu x - 5 < 0 => x < 5 thì (1) trở thành: -(x-5)=7-2x <=> -x+5=7-2x <=> x=2 (nhận)

Vậy x = 2

\(\sqrt{x-2}+\sqrt{2-x}=0\)

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

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

\(\Leftrightarrow2\sqrt{4x-x^2-4}=0\)

\(\Leftrightarrow\left(\sqrt{4x-x^2-4}\right)^2=0\)

\(\Leftrightarrow4x-x^2-4=0\)

giải phương trình bình thường

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

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

\(\Leftrightarrow x^2+x+1=x^2+4x+4\)

\(\Leftrightarrow-3x=3\)

\(\Leftrightarrow x=-1\)

Vậy x = -1

Cảm ơn bạn nha

Từ GT <->   \(x+y+z=2\sqrt{x}+4\sqrt{y}+6\sqrt{z}-14\)

          <>     \(\left(x-2\sqrt{x}+1\right)\)+ \(\left(y-4\sqrt{y}+4\right)+\left(z-6\sqrt{z}+9\right)\)\(=0\)

        <>        \(\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-2\right)^2+\left(\sqrt{z}-3\right)^2=0\)

vì \(\left(\sqrt{x}-1\right)^2\ge0\forall x>0\).......................................................................

đến đây tự làm tiếp nhé

\(a,\sqrt{x}=7\)

\(\Rightarrow\sqrt{x}=\sqrt{49}\)

\(\Rightarrow x=49\)

\(b,\sqrt{x^3}=0\)

\(\Rightarrow x^3=0\)

\(\Rightarrow x=0\)

a) \(\sqrt{x}=7\Rightarrow x=49\)

b) \(\sqrt{x^3}=0\Rightarrow x=0\)