\(14\cdot7^{2021}=35\cdot7^{2021}-3\cdot49^x\)

\(\Rightarrow35\cdot7^{2021}-3\cdot49^x=14\cdot7^{2021}\)

\(\Rightarrow5\cdot7\cdot7^{2021}-3\cdot\left(7^2\right)^x=2\cdot7\cdot7^{2021}\)

\(\Rightarrow5\cdot7^{2022}-3\cdot7^{2x}=2\cdot7^{2022}\)

\(\Rightarrow3\cdot7^{2x}=5\cdot7^{2022}-2\cdot7^{2022}\)

\(\Rightarrow3\cdot7^{2x}=\left(5-2\right)\cdot7^{2022}\)

\(\Rightarrow3\cdot7^{2x}=3\cdot7^{2022}\)

\(\Rightarrow7^{2x}=7^{2022}\)

\(\Rightarrow2x=2022\)

\(\Rightarrow x=2022:2\)

\(\Rightarrow x=1011\)

Vậy \(x=1011\).

\(a,\Leftrightarrow\left(x-9\right)^2-2\left(x-9\right)+1=0\\ \Leftrightarrow\left(x-9-1\right)^2=0\Leftrightarrow x=10\\ b,Sửa:49x^2-14x\sqrt{5}+5=0\\ \Leftrightarrow\left(7x-\sqrt{5}\right)^2=0\Leftrightarrow x=\dfrac{\sqrt{5}}{7}\)

\(\Leftrightarrow x\left(x^6-14x^4+49x^2-36\right)=0\)

\(\Leftrightarrow x\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=1\\x^2=4\\x^2=9\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\\x=\pm2\\x=\pm3\end{matrix}\right.\)

ĐKXĐ: \(x\ge2\)

Từ pt đã cho suy ra:

\(7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8\)

⇒ \(2\sqrt{x-2}=8\) ⇒ \(x=18\)

49. x³-x=0

<=> 49. x(x²-1)=0

<=> 49. x(x²-1²)=0

<=> 49.x(x-1)(x+1)=0

=> x=0 hoặc x-1=0 hoặc x+1=0

=> x=0 hoặc x=1 hoặc x= -1

49. x³-x=0

<=> 49. x(x²-1)=0

<=> 49. x(x²-1²)=0

<=> 49.x(x-1)(x+1)=0

=> x=0 hoặc x-1=0 hoặc x+1=0

=> x=0 hoặc x=1 hoặc x= -1

Giải:

\(\sqrt{49x-98}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8\)

ĐKXĐ: \(x-2\ge0\Leftrightarrow x\ge2\)

\(7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8\)

\(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

\(\Leftrightarrow2\sqrt{x-2}=8\)

\(\Leftrightarrow\sqrt{x-2}=4\)

\(\Leftrightarrow x-2=16\)

\(\Leftrightarrow x=18\) (thỏa mãn)

Vậy ...

a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$

$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$

$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$

Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}=\frac{12}{5}$

$\Leftrightarrow x=5,76$ (thỏa mãn)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 2\sqrt{x-2}=8$

$\Leftrightarrow \sqrt{x-2}=4$

$\Leftrightarrow x=4^2+2=18$ (tm)

b. ĐKXĐ: $x^2\geq 5$

PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$

$\Leftrightarrow \sqrt{x^2-5}=0$

$\Leftrightarrow x=\pm \sqrt{5}$