Lời giải:
PT(1)\(\Rightarrow y^2=x-1\Rightarrow x\geq 1\)

Thay $y^2=x-1$ vào PT(2) ta có:

\(x-1+x(x+1)(x+2)(x+3)=121\)

\(\Leftrightarrow x+[x(x+3)][(x+1)(x+2)]=122\)

\(\Leftrightarrow x+(x^2+3x)(x^2+3x+2)=122\)

\(\Leftrightarrow x+(x^2+3x)^2+2(x^2+3x)=122\)

\(\Leftrightarrow x^4+6x^3+11x^2+7x-122=0\)

\(\Leftrightarrow x^3(x-2)+8x^2(x-2)+27x(x-2)+61(x-2)=0\)

\(\Leftrightarrow (x-2)(x^3+8x^2+27x+61)=0\)

Với $x\geq 1$ thì \(x^3+8x^2+27x+61>0\)

Do đó \(x-2=0\Leftrightarrow x=2\) là nghiệm duy nhất.

Thay vào PT đầu tiên \(y^2=x-1=2-1=1\Rightarrow y=\pm 1\)

Vậy \((x,y)=(2,\pm 1)\)

\(\left(x-\sqrt{121}\right).\left(x^2+1\right)=\)\(0\)

\(\Rightarrow\)\(\orbr{\begin{cases}x-\sqrt{121}=0\\x^2+1=0\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=\\x=\end{cases}}\)

bn tự làm tiếp nha

\(\left(4x-1\right)^2-x\left(3-x\right)=121\)

\(\Leftrightarrow16x^2-8x+1-3x+x^2=121\)

\(\Leftrightarrow17x^2-11x-120=0\)

\(\Leftrightarrow17x^2-51x+40x-120=0\)

\(\Leftrightarrow\left(17x^2-51x\right)+\left(40x-120\right)=0\)

\(\Leftrightarrow17x\left(x-3\right)+40\left(x-3\right)=0\)

\(\Leftrightarrow\left(17x+40\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}17x+40=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-40}{17}\\x=3\end{matrix}\right.\)

\(5x^2-x=18\)

\(\Leftrightarrow5x^2-x-18=0\)

\(\Leftrightarrow5x^2-10x+9x-18=0\)

\(\Leftrightarrow\left(5x^2-10x\right)+\left(9x-18\right)=0\)

\(\Leftrightarrow5x\left(x-2\right)+9\left(x-2\right)=0\)

\(\Leftrightarrow\left(5x+9\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+9=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9}{5}\\x=2\end{matrix}\right.\)

Lời giải:

$(3x+2)^2=121=11^2=(-11)^2$

$\Rightarrow 3x+2=11$ hoặc $3x+2=-11$

$\Rightarrow x=3$ hoặc $x=\frac{-13}{3}$

Vì $x$ là số tự nhiên nên $x=3$

\(\left(1\dfrac{3}{4}-\dfrac{4}{6}\right):\left(1\dfrac{1}{5}+2\dfrac{2}{5}+\dfrac{1}{5}\right)< x< 1\dfrac{1}{5}.1\dfrac{1}{4}+3\dfrac{2}{11}:2\dfrac{3}{121}\)

c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)

=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)

=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)

=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)

d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)

=> \(\left(2x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)

=> 2x-1 = \(\dfrac{-2}{3}\)

=> x= \(\dfrac{1}{6}\)

a. \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

=> \(\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)

=> x + \(\dfrac{1}{2}\) = \(\dfrac{1}{4}\) hoặc \(x+\dfrac{1}{2}=\dfrac{-1}{4}\)

=> x = \(\dfrac{-1}{4}\) hoặc \(x=\dfrac{-3}{4}\)

Đặt \(x^2+2x=t\left(đk:t\ge-1\right).\)

\(\Rightarrow\left(11-t\right)\left(11+t\right)=121\Leftrightarrow121-t^2=121\Leftrightarrow t=0\)

\(\Rightarrow x^2+2x=0\Leftrightarrow\orbr{\begin{cases}-2\\0\end{cases}}\)

Thực ra bài này đặt cho dễ nhìn, ta hoàn toàn có thể ssuy ra mà ko cần đặt ha!

= -2 hoặc 0 

~ Hok tốt ~

#JH

bn ghi như rứa mk ko hiểu

Tìm số tự nhiên xx thỏa mãn: \left(3.x+2\right)^{2}=121(3.x+2)2=121