a. Ta có: \(x^2-10x+26+y^2+2y=0\Leftrightarrow\left(x^2-10x+25\right)+\left(y^2+2y+1\right)=0\\ \)

\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\Rightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}}\)

b. \(\left(2x+5\right)^2-\left(x-7\right)^2=0\Leftrightarrow\left(2x+5+x-7\right).\left(2x+5-x+7\right)=0\)

\(\Leftrightarrow\left(3x-2\right).\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-12\end{cases}}}\)

c. \(25.\left(x-3\right)^2=49.\left(1-2x\right)^2\Leftrightarrow\left(5x-15\right)^2=\left(7-14x\right)^2\Leftrightarrow\left(5x-15\right)^2-\left(7-14x\right)^2=0\)

\(\Leftrightarrow\left(5x-15-7+14x\right).\left(5x-15+7-14x\right)=0\Leftrightarrow\left(19x-22\right).\left(-9x-8\right)=0\)

\(\Leftrightarrow\left(19x-22\right).\left(9x+8\right)=0\Leftrightarrow\orbr{\begin{cases}19x-22=0\\9x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{22}{19}\\x=-\frac{8}{9}\end{cases}}}\)

d. \(\left(x+2\right)^2=\left(3x-5\right)^2\Leftrightarrow\left(x+2\right)^2-\left(3x-5\right)^2=0\Leftrightarrow\left(x+2+3x-5\right).\left(x+3-3x+5\right)=0\)

\(\Leftrightarrow\left(4x-3\right).\left(8-2x\right)=0\Leftrightarrow\orbr{\begin{cases}4x-3=0\\8-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=4\end{cases}}}\)

e. \(x^2-2x+1=16\Leftrightarrow\left(x-1\right)^2-16=0\Leftrightarrow\left(x-1-4\right).\left(x-1+4\right)=0\)

\(\Leftrightarrow\left(x-5\right).\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)

Cảm ơn bn rất nhìu nha!!!^-^!!!

Bn ơi cho mik hỏi cái này đc ko ở câu a hai số cộng vs nhau bằng 0 thì một trong hai số đó là băng 0 à bn . Hi hi mik lại nghĩ là hai số nhân vs nhau bằng 0 thì 1 trong hai số đó ms bằng 0 chứ!!! Dù sao thì cx cảm ơn bn nhìu nà!!! Chúc bn học tốt nha!!!

C

Ta có:

a)ps tối giản

b) 9/21= 3/7

c) 6/16=3/8

d)8/14=4/7

=> ps bé hơn 3/7 là: d)8/14

c nha

\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)-3x\left(1-x\right)\)

\(=x^3-9x^2+27x-27-\left(x^3-27\right)-3x+3x^2\)

\(=x^3-9x^2+27x-27-x^3+27-3x+3x^2\)

\(=24x-6x^2\)

Hình như đề có chỗ sai sót ở đâu đó bạn .

vãi ò ông ngx thành đạt chép sai đầu bài r (x-1)³ cchchuchưchứchứ kkoko pphphaphaiphảiphải (x-3)³³

KHO THE

\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)

\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)

\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)

a) Ta có: \(\dfrac{7\cdot25}{14\cdot10}\)

\(=\dfrac{7\cdot5\cdot5}{7\cdot2\cdot2\cdot5}\)

\(=\dfrac{5}{4}\)

b) Ta có: \(\dfrac{24\cdot15-14\cdot9}{36\cdot12}\)

\(=\dfrac{9\cdot8\cdot5-14\cdot9}{36\cdot12}\)

\(=\dfrac{9\cdot\left(8\cdot5-14\right)}{9\cdot4\cdot12}\)

\(=\dfrac{40-14}{4\cdot12}\)

\(=\dfrac{13}{24}\)

a) \(\dfrac{7.25}{14.10}\)

= \(\dfrac{7.5.5}{7.2.5.2}\)

= \(\dfrac{5}{4}\)

c) \(\dfrac{1^3.3^2}{6^3.5}\)

= \(\dfrac{3^2}{\left(2.3\right)^3.5}\)

= \(\dfrac{1}{2^3.3.5}\)

= \(\dfrac{1}{120}\)

\(\dfrac{3}{4}:\left(2\dfrac{4}{9}\right)-\left|-3x+2\dfrac{2}{3}\right|=\dfrac{3}{4}\)

\(\Leftrightarrow\left|-3x+\dfrac{8}{3}\right|=\dfrac{3}{4}-\dfrac{3}{4}\cdot\dfrac{9}{22}\)

\(\Leftrightarrow\left|3x-\dfrac{8}{3}\right|=\dfrac{3}{4}-\dfrac{27}{88}=\dfrac{39}{88}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{8}{3}=\dfrac{39}{88}\\3x-\dfrac{8}{3}=-\dfrac{39}{88}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{821}{792}\\x=\dfrac{587}{792}\end{matrix}\right.\)

Trả lời:

\(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy x = -1/15; x = 1/3

Cảm ơn mn nhìu lém

Trả lời:

\(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left(\dfrac{2}{5}-3x\right)^2=\left(\dfrac{3}{5}\right)^2\) hoặc \(\left(\dfrac{2}{5}-3x\right)^2=\left(-\dfrac{3}{5}\right)^2\)

\(\Leftrightarrow\dfrac{2}{5}-3x=\dfrac{3}{5}\) hoặc \(\dfrac{2}{5}-3x=-\dfrac{3}{5}\)

\(\Leftrightarrow3x=\dfrac{2}{5}-\dfrac{3}{5}\) hoặc \(3x=\dfrac{2}{5}-\left(-\dfrac{3}{5}\right)\)

\(\Leftrightarrow3x=-\dfrac{1}{5}\) hoặc \(3x=1\)

\(\Leftrightarrow x=-\dfrac{1}{5}:3\) hoặc \(x=1:3\)

\(\Leftrightarrow x=-\dfrac{1}{15}\) hoặc \(x=\dfrac{1}{3}\)

Vậy ...