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

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

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

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

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

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

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

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

\(\Rightarrow2x=2022\)

\(\Rightarrow x=2022:2\)

\(\Rightarrow x=1011\)

Vậy \(x=1011\).

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

\(x=\frac{}{}\frac{\sqrt5+2}{7}hoặc\frac{\sqrt5-2}{7}\)

Ta có: \(49x^2-14\sqrt5x+6=5\)

=>\(49x^2-14\sqrt5x+1=0\)

=>\(\left(7x\right)^2-2\cdot7x\cdot\sqrt5+5-4=0\)

=>\(\left(7x-\sqrt5\right)^2=4=2^2\)

=>\(\left[\begin{array}{l}7x-\sqrt5=2\\ 7x-\sqrt5=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}7x=2+\sqrt5\\ 7x=\sqrt5-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{2+\sqrt5}{7}\\ x=\frac{\sqrt5-2}{7}\end{array}\right.\)

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

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

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

Cho em hỏi Nguyễn Việt Lâm có phải gv không ạ?

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

Đặt A=3+15+35+...+9215

\(=1\cdot3+3\cdot5+\cdots+95\cdot97\)

\(=1\left(1+2\right)+3\left(3+2\right)+\ldots+95\left(95+2\right)\)

\(=\left(1^2+3^2+\cdots+95^2\right)+2\left(1+3+\cdots+95\right)\)

Đặt \(B=1^2+3^2+\cdots+95^2\)

\(=1^2+2^2+3^2+4^2+\cdots+96^2-\left(2^2+4^2+\cdots+96^2\right)\)

\(=\frac{96\left(96+1\right)\left(2\cdot96+1\right)}{6}-2^2\left(1^2+2^2+\cdots+48^2\right)\)

\(=16\cdot97\cdot193-4\cdot\frac{48\left(48+1\right)\left(2\cdot48+1\right)}{6}\)

\(=16\cdot97\cdot193-\frac23\cdot48\cdot49\cdot97=16\cdot97\cdot193-32\cdot49\cdot97\)

\(=16\cdot97\left(193-2\cdot49\right)=16\cdot97\cdot95=147440\)

Số số hạng trong dãy số 1;3;...;95 là:

(95-1):2+1=94:2+1=47+1=48(số)

Ta có: |x+3|+|x+15|+...+|x+9215|=49x

=>49x>=0

=>x>=0

Phương trình sẽ trở thành:

x+3+x+15+...+x+9215=49x

=>48x+147440=49x

=>x=147440(nhận)

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 ...