\(\left(3x+1\right)^2=\left(15-7x\right)^2\)

\(\Leftrightarrow\left(3x+1\right)^2-\left(15-7x\right)^2=0\)

\(\Leftrightarrow\left(3x+1+15-7x\right)\left(3x+1-15+7x\right)=0\)

\(\Leftrightarrow\left(16-4x\right)\left(10x-14\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}16-4x=0\\10x-14=0\end{cases}\Leftrightarrow\orbr{\begin{cases}4x=16\\10x=14\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=\frac{7}{5}\end{cases}}}\)

Vậy ...............

(x - 2)(x + 3) = 50

=> x² + 3x - 2x - 6 - 50 = 0

=> x² + x - 56 = 0

=> x² + 8x - 7x - 56 = 0

=> x(x + 8) - 7(x + 8) = 0

=> (x - 7)(x + 8)  = 0

=> \(\orbr{\begin{cases}x-7=0\\x+8=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=7\\x=-8\end{cases}}\)

\(\left(x-2\right)\left(x+3\right)=50\)

\(x^2+x-6=50\)

\(x^2+x-56=0\)

\(x^2-7x+8x-56=0\)

\(x\cdot\left(x-7\right)-8\cdot\left(x-7\right)=0\)

\(\left(x-7\right)\cdot\left(x-8\right)=0\)

\(\orbr{\begin{cases}x-7=0\\x-8=0\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}}\)

2xy + x² + y² = ( x + y )² = ( x+y ) ( x + y )

( 25z -15n )² = ( 25z - 15n ) ( 25z -15n )

\(\left(A+B\right)^2=A^2+2AB+B^2\)

\(\left(A-B\right)^2=A^2-2AB+B^2\)

\(2xy+x^2+y^2=x^2+2.x.y+y^2=\left(x+y\right)^2\)\(\left(25z-15n\right)^2=\left(25n\right)^2-2.25z.15n+\left(15n\right)^2=625z^2-375zn+225n^2\)

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=2+2\left(ab+bc+ac\right)\)

=> \(0=2+2\left(ab+bc+ac\right)\)=> \(ab+bc+ca=-1\)

=> \(\left(ab+bc+ac\right)^2=1\)

Mà \(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+a^2bc+abc^2\right)\)

                                             \(=a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+a^2c^2\)

=> \(a^2b^2+b^2c^2+c^2a^2=1\)

Mặt khác : \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

=> \(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

                                             \(=4-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

=> \(a^4+b^4+c^4=4-2=2\)

Có \(B=n^4-27n^2+121\)

\(=n^4+22n^2+121-49n^2\)

\(=\left(n^2+11\right)^2-\left(7n\right)^2\)

\(=\left(n^2+11-7n\right)\cdot\left(n^2+11+7n\right)\)

Vì \(n\in N\)nên \(n^2+7n+11>11\)

Nếu \(n^2-7n+11< 0\Rightarrow B< 0\left(loại\right)\)

Nếu \(n^2-7n+11=0\Rightarrow B=0\left(loại\right)\)

Nếu \(n^2-7n+11>1\)(loại vì B là tích của 2 số nguyên dương > 1 nên ko là số nguyên tố)

Vậy nên \(n^2-7n+11=1\)

\(\Leftrightarrow n^2-7n+10=0\)

\(\Leftrightarrow n^2-2n-5n+10=0\)

\(\Leftrightarrow\left(n-2\right)\cdot\left(n-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}n-2=0\\n-5=0\end{cases}\Rightarrow\orbr{\begin{cases}n=2\\n=5\end{cases}}}\)

Vậy.............

ấn vào đúng cho mk đi mk ân cho bạn ok

Bài 1: Chỉ cần chú ý đẳng thức \(a^5+b^5=\left(a^2+b^2\right)\left(a^3+b^3\right)-a^2b^2\left(a+b\right)\) là ok! 

Làm như sau: Từ \(x^2+\frac{1}{x^2}=14\Rightarrow x^2+2.x.\frac{1}{x}+\frac{1}{x^2}=16\)

\(\Rightarrow\left(x+\frac{1}{x}\right)^2=16\). Do \(x>0\Rightarrow x+\frac{1}{x}>0\Rightarrow x+\frac{1}{x}=4\)

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

\(=14\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)\)

\(=14\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}-1\right)-4\)

\(=14.4.\left(14-1\right)-4=724\) là một số nguyên (đpcm)

P/s: Lâu ko làm nên cũng ko chắc đâu nhé!