a) \(8 - \left( {x - 15} \right) = 2.\left( {3 - 2x} \right)\) 

\(8 - x + 15 = 6 - 4x\)

\( - x + 4x = 6 - 8 - 15\)

\(3x =  - 17\)

\(x = \left( { - 17} \right):3\)

\(x = \dfrac{{ - 17}}{3}\)

Vậy nghiệm của phương trình là \(x = \dfrac{{ - 17}}{3}\).

b) \( - 6\left( {1,5 - 2u} \right) = 3\left( { - 15 + 2u} \right)\)

\( - 9 + 12u =  - 45 + 6u\)

\(12u - 6u =  - 45 + 9\)

\(u = \left( { - 36} \right):6\)

\(6u =  - 36\)

\(u =  - 6\)

Vậy nghiệm của phương trình là \(u =  - 6\).

c) \({\left( {x + 3} \right)^2} - x\left( {x + 4} \right) = 13\)

\(\left( {{x^2} + 6x + 9} \right) - \left( {{x^2} + 4x} \right) = 13\)

\({x^2} + 6x + 9 - {x^2} - 4x = 13\)

\(\left( {{x^2} - {x^2}} \right) + \left( {6x - 4x} \right) = 13 - 9\)

\(2x = 4\)

\(x = 4:2\)

\(x = 2\)

Vậy nghiệm của phương trình là \(x = 2\).

d) \(\left( {y + 5} \right)\left( {y - 5} \right) - {\left( {y - 2} \right)^2} = 5\)

\(\left( {{y^2} - 25} \right) - \left( {{y^2} - 4y + 4} \right) = 5\)

\({y^2} - 25 - {y^2} + 4y - 4 = 5\)

\(\left( {{y^2} - {y^2}} \right) + 4y = 5 + 4 + 25\)

\(4y = 34\)

\(y = 34:4\)

\(y = \dfrac{{17}}{2}\)

Vậy nghiệm của phương trình là \(y = \dfrac{{17}}{2}\).

a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)

=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75

=>x=7; y=5

b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)

=>4x+9y=8 và -8x+3y=5

=>x=-1/4; y=1

c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)

=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5

=>2x-3y=-5,5 và 3x-2y=-4,5

=>x=-1/2; y=3/2

e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)

=>\(x=\sqrt{2};y=\sqrt{3}\)

\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{1}{x-7}+\dfrac{1}{x-7}-\dfrac{1}{x-13}+\dfrac{1}{x-13}-\dfrac{1}{x-28}-\dfrac{1}{x-28}=\dfrac{-5}{2}\)

\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{2}{x-28}=-\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x-28-2x+8}{\left(x-4\right)\left(x-28\right)}=\dfrac{-5}{2}\)

\(\Leftrightarrow-5\left(x^2-32x+112\right)=2\left(-x-20\right)\)

\(\Leftrightarrow-5x^2+160x-560=-2x-40\)

\(\Leftrightarrow-5x^2+162x-520=0\)

\(\text{Δ}=162^2-4\cdot\left(-5\right)\cdot\left(-520\right)=15844\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{162-2\sqrt{3961}}{10}\\x_2=\dfrac{162+2\sqrt{3961}}{10}\end{matrix}\right.\)

Điều kiện xác định : \(x,y,z\ge0\)

Đặt \(a=\sqrt{x}-13\) , \(b=\sqrt{y}-14\) , \(c=\sqrt{z}-15\)

Ta có hệ : \(\hept{\begin{cases}ab=2\\bc=6\\ac=3\end{cases}}\). Nhân các pt theo vế : \(\left(abc\right)^2=36\Leftrightarrow\orbr{\begin{cases}abc=6\\abc=-6\end{cases}}\)

TH1. Nếu abc = 6 thì kết hợp với mỗi pt ta được : \(\hept{\begin{cases}c=3\\b=2\\a=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=196\\y=256\\z=324\end{cases}}\)

TH2. Nếu \(abc=-6\) thì tương tự ta được \(\hept{\begin{cases}a=-1\\b=-2\\c=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=144\\y=144\\z=144\end{cases}}\)

Vậy ................................................

CHIU THOI

K NHA @@@@@@@ Nguyễn Phúc Lộc 

a) \(\left(-12x^{13}y^{15}+6x^{10}y^{14}\right):\left(-3x^{10}y^{14}\right)\)

\(=-12x^{13}y^{15}:-3x^{10}y^{14}+6x^{10}y^{14}:-3x^{10}y^{14}\)

\(=4x^3y-2\)

b) \(\left(x-y\right)\left(x^2-2x+y\right)-x^3+x^2y\)

\(=x^3-2x^2+xy-x^2y+2xy-y^2-x^3+x^2y\)

\(=-2x^2+3xy-y^2\) 

a) \(-12x^{13}\)\(y^{15}\)+\(6x^{10}\)\(y^{14}\):\(-3x^{10}\)\(y^{14}\)

=\(-12x\)\(^{13}\)\(y^{15}\)\(:\)\(-3x^{10}y^{14}\)\(+6x^{10}y^{14}:-3x^{10}y^{14}\)

\(=4x^3y-2\)

b)\(=\left(x-y\right)x^2-2x+y-x^3+x^2y\)

\(=x^3-x^2y-2x+y-x^3+x^2y\)

\(=-2x+y\)

A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)

\(=\frac{1}{x+3}-\frac{1}{x+34}\)

\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)

\(\Rightarrow x=31\)

Vậy, x = 31 

Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với    \(x,k\inℝ;x\ne0;x\ne-k\)

Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)

a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2+y^2\right)+\left(x^2+y^2-4\right)\left(y+2\right)=0\\x^2+y^2+\left(x+y-2\right)\left(y+2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x^2+y^2-4\right)\left(y+2\right)=-x\left(x^2+y^2\right)\\-\left(x^2+y^2\right)=\left(x+y-2\right)\left(y+2\right)\end{matrix}\right.\)

\(\Rightarrow\left(x^2+y^2-4\right)\left(y+2\right)=x\left(x+y-2\right)\left(y+2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}y+2=0\left(\text{không thỏa mãn}\right)\\x^2+y^2-4=x\left(x+y-2\right)\end{matrix}\right.\) 

\(\Rightarrow x^2+y^2-4=x^2+x\left(y-2\right)\)

\(\Leftrightarrow\left(y+2\right)\left(y-2\right)=x\left(y-2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2\\x=y+2\end{matrix}\right.\)

Thế vào pt dưới:

\(\Rightarrow\left[{}\begin{matrix}x^2+8+2x+2x-4=0\\\left(y+2\right)^2+2y^2+y\left(y+2\right)+2\left(y+2\right)-4=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

Câu b chắc chắn đề sai, nhìn 2 vế pt đầu đều có \(x^2\) thì chúng sẽ rút gọn, không ai cho đề như thế hết

Mk sửa lại đề rồi. Bạn giúp mk giải vs