a: Đặt |x-6|=a, |y+1|=b

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}2a+3b=5\\5a-4b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

=>|x-6|=1 và |y+1|=1

\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)

b: Đặt |x+y|=a, |x-y|=b

Theo đề, ta có: \(\left\{{}\begin{matrix}2a-b=19\\3a+2b=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{55}{7}\\b=-\dfrac{23}{7}\left(loại\right)\end{matrix}\right.\)

=>HPTVN

c: Đặt |x+y|=a, |x-y|=b

Theo đề ta có: \(\left\{{}\begin{matrix}4a+3b=8\\3a-5b=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=0\end{matrix}\right.\)

=>|x+y|=2 và x=y

=>|2x|=2 và x=y

=>x=y=1 hoặc x=y=-1

\(a,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)

\(=2x^2+2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3\left(x^2+y^2\right)\)\(b,\left(5x-1\right)+2\left(1-5x\right)\left(4x+5\right)+\left(5x+4\right)\)\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2=25\)

c)\(Q=\left(x-y\right)^3+\left(x+y\right)^3+\left(x-y\right)^3-3xy\left(x+y\right)\)

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

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

d)\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(2P=5^{32}-1\Rightarrow P=\dfrac{5^{32}-1}{2}\)

a) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)

\(\Leftrightarrow2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy+y^2\)

\(\Leftrightarrow2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2\)

\(\Leftrightarrow4x^2\)

Bài 1:

a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)

\(=10-10x=10(1-x)\)

b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)

\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)

\(=-7x^2+7x=7x(1-x)\)

c)

\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)

\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)

\(=\left\{3-x-5[9x-2]\right\}(-2x)\)

\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)

Bài 2:

a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)

\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)

\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)

b)

\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)

\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)

\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)

\(2x^2+3(x^2-1)=5x(x+1)\)

Bài 2:

c) \(2x(5-3x)+2x(3x-5)-3(x-7)=3\)

\(\Leftrightarrow 2x(5-3x)-2x(5-3x)-3(x-7)=3\)

\(\Leftrightarrow -3(x-7)=3\)

\(\Leftrightarrow x-7=-1\Rightarrow x=6\)

d)

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

\(\Leftrightarrow 3x^2+3x-(2x^2+4x)+x+1=0\)

\(\Leftrightarrow x^2+1=0\)

Vô lý vì \(x^2+1\geq 0+1=1>0\) với mọi $x$

Vậy không tồn tại $x$ thỏa mãn.

a. \(2x\left(x-5\right)-x\left(2x+3\right)=26\Rightarrow2x^2-10x-2x^2-3x=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

b. \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\frac{5}{2}\)

\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3=\frac{5}{2}\)\(\Rightarrow2y=\frac{7}{2}\Rightarrow y=\frac{7}{4}\)

c. \(2x^2+3\left(x+1\right)\left(x-1\right)=5x^2+5x\Rightarrow5x^2-3=5x^2+5x\)

\(\Rightarrow x=-\frac{3}{5}\)

cảm ơn bạn nhiều nhé 

kb vs mình đi 

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

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

=>\(y'=2x^2-3+\left(x+2\right)\cdot2x\)

\(\Leftrightarrow y'=2x^2-3+2x^2+4x=4x^2+4x-3\)

b: \(y=\left(x-1\right)^2\left(x+2\right)\)

=>\(y=\left(x^2-2x+1\right)\left(x+2\right)\)

=>\(y'=\left(x^2-2x+1\right)'\left(x+2\right)+\left(x^2-2x+1\right)\left(x+2\right)'\)

=>\(y'=\left(2x-2\right)\left(x+2\right)+\left(x^2-2x+1\right)\)

=>\(y'=2x^2+4x-2x-4+x^2-2x+1\)

=>\(y'=3x^2-3\)

c: \(y=\left(x^2-1\right)\left(2x+1\right)\)

=>\(y'=\left(x^2-1\right)'\left(2x+1\right)+\left(x^2-1\right)\left(2x+1\right)'\)

=>\(y'=2x\left(2x+1\right)+2\left(x^2-1\right)\)

=>\(y'=4x^2+2x+2x^2-2=6x^2+2x-2\)

d: \(y=\left(x+2\right)\left(2x^2-5\right)\)

=>\(y'=\left(x+2\right)'\left(2x^2-5\right)+\left(x+2\right)\left(2x^2-5\right)'\)

=>\(y'=2x^2-5+2x\left(x+2\right)=4x^2+4x-5\)

a) \(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}xy+5x+3y+15=xy+8x+y+8\\10xy+14x-15y-21=10xy+10x-12y-12\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-3x+2y=-7\\4x-3y=9\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-9x+6y=-21\\8x-6y=18\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-x=-3\\8x-6y=18\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=3\\8.3-6y=18\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm (x;y)=(3;1)

b) ĐKXĐ:\(\left\{{}\begin{matrix}2y-5\ne0\\3y-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\ne\dfrac{5}{2}\\y\ne\dfrac{4}{3}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{2x-3}{2y-5}=\dfrac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}\left(2x-3\right)\left(3y-4\right)=\left(3x+1\right)\left(2y-5\right)\\2x-6-3y-6=-16\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}6xy-8x-9y+12=6xy-15x+2y-5\\2x-3y=-4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}7x-11y=-17\\2x-3y=-4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}14x-22y=-34\\14x-21y=-28\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}14x-22y=-34\\-y=-6\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}14x-22.6=-34\\y=6\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=7\left(TM\right)\\y=6\left(TM\right)\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm (x;y)=(7;6)

a: \(y=\left(5x-10\right)^4\)

=>\(y'=4\cdot\left(5x-10\right)'\cdot\left(5x-10\right)^3\)

\(=4\cdot5\cdot\left(5x-10\right)^3=20\left(5x-10\right)^3\)

Đặt y'>0

=>\(20\left(5x-10\right)^3>0\)

=>\(\left(5x-10\right)^3>0\)

=>5x-10>0

=>x>2

Đặt y'<0

=>\(20\left(5x-10\right)^3< 0\)

=>\(\left(5x-10\right)^3< 0\)

=>5x-10<0

=>x<2

Vậy: hàm số đồng biến trên \(\left(2;+\infty\right)\)

Hàm số nghịch biến trên \(\left(-\infty;2\right)\)

c: \(y=\left(x^3-1\right)^3\)

=>\(y'=3\left(x^3-1\right)'\cdot\left(x^3-1\right)^2\)

\(=9x^2\left(x^3-1\right)^2>=0\forall x\)

=>Hàm số luôn đồng biến trên R

d: \(y=\left(x^2-1\right)\left(x+2\right)\)

=>\(y'=\left(x^2-1\right)'\left(x+2\right)+\left(x^2-1\right)\left(x+2\right)'\)

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

\(=2x^2+4x+x^2-1=3x^2+4x-1\)

Đặt y'>0

=>\(3x^2+4x-1>0\)

=>\(\left[{}\begin{matrix}x< \dfrac{-2-\sqrt{7}}{3}\\x>\dfrac{-2+\sqrt{7}}{3}\end{matrix}\right.\)

Đặt y'<0

=>\(3x^2+4x-1< 0\)

=>\(\dfrac{-2-\sqrt{7}}{3}< x< \dfrac{-2+\sqrt{7}}{3}\)

Vậy: Hàm số đồng biến trên các khoảng \(\left(-\infty;\dfrac{-2-\sqrt{7}}{3}\right);\left(\dfrac{-2+\sqrt{7}}{3};+\infty\right)\)

Hàm số nghịch biến trên khoảng \(\left(\dfrac{-2-\sqrt{7}}{3};\dfrac{-2+\sqrt{7}}{3}\right)\)

b: \(y=\left(-x-1\right)\left(x+2\right)^4\)

=>\(y'=\left(-x-1\right)'\left(x+2\right)^4+\left(-x-1\right)\left[\left(x+2\right)^4\right]'\)

\(=-\left(x+2\right)^4+\left(-x-1\right)\cdot4\left(x+2\right)'\left(x+2\right)^3\)

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

\(=-\left(x+2\right)^3\left[\left(x+2\right)+4\left(x+1\right)\right]\)

\(=-\left(x+2\right)^2\cdot\left(x+2\right)\left(5x+6\right)\)

Đặt y'<0

=>\(-\left(x+2\right)^2\left(x+2\right)\left(5x+6\right)< 0\)

=>(x+2)(5x+6)>0

TH1: \(\left\{{}\begin{matrix}x+2>0\\5x+6>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>-2\\x>-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x>-\dfrac{6}{5}\)

TH2: \(\left\{{}\begin{matrix}x+2< 0\\5x+6< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< -2\\x< -\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x< -2\)

Đặt y'>0

=>(x+2)(5x+6)<0

TH1: \(\left\{{}\begin{matrix}x+2>0\\5x+6< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>-2\\x< -\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow-2< x< -\dfrac{6}{5}\)

TH2: \(\left\{{}\begin{matrix}x+2< 0\\5x+6>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< -2\\x>-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy: HSĐB trên các khoảng \(\left(-\infty;-2\right);\left(-\dfrac{6}{5};+\infty\right)\)

HSNB trên khoảng \(\left(-2;-\dfrac{6}{5}\right)\)

\(a,\left(x+3\right)\left(y+2\right)=1\)

=> x+3 và y+2 thuộc UC(1)={1; -1}

Vậy x=-2; y=-4

       x=-1; y=-4

Câu sau tương tự

\(a,\left(x+3\right)\left(y+2\right)=1\)

Th1 : \(\hept{\begin{cases}x+3=1\\y+2=1\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\y=-1\end{cases}}}\)

Th2 : \(\hept{\begin{cases}x+3=-1\\y+2=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=-3\end{cases}}}\)

KL : \(\left\{\left(x=-2;y=-1\right);\left(x=-4;y=-3\right)\right\}\)

\(d,3x+4y-xy=16\)

\(=3x-xy+4y-12=4\)

\(\Rightarrow-x\left(y-3\right)+4\left(y-3\right)=4\)

\(\Rightarrow\left(y-3\right)\left(4-x\right)=4\)

Chia các trường hợp như câu a của chị ra em nhé

vì \(\left(x+3\right)\left(y+2\right)=1\)

\(\Rightarrow\left(x+3\right)\text{và}\left(y+2\right)=1\)

\(x+3=1\)                                   \(y+2=1\)

\(x=1-3\)                                  \(y=1-2\)

\(x=-2\)                                      \(y=-1\)