...

\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)

\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)

đề ở câu đầu:

x-3/0,2=2x-1/-0,5

3x-1/2x+3=3x+2/2x-1

a: =>-0,5x+1,5=0,4x-0,2

=>-0,9x=-1,7

=>x=17/9

3x-1/2x+3=3x+2/2x-1

=>6x^2-3x-2x+1=6x^2+4x+9x+6

=>-5x+1=13x+6

=>-8x=5

=>x=-5/8

b: \(\Leftrightarrow\left(4x-1\right)\left(-x+7\right)=\left(4x+5\right)\left(-x-2\right)\)

=>\(-4x^2+28x+x-7=-4x^2-8x-5x-10\)

=>29x-7=-13x-10

=>42x=-3

=>x=-1/14

c: =>7x=5y và 2x-y=15

=>7x-5y=0 và 2x-y=15

=>x=25; y=35

khó quá à

Chỉ là cảm thấy dài 

TL:

\(a,\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)

\(=6x^2+23x-55-6x^2-23x-21\)

\(=-76\)

Mấy câu kia tương tự !

\(4.\left(3x+y\right)^2+\left(x+y\right)^2\) 

\(=3x^2+6xy+y^2+x^2-2xy+y^2\) 

\(=9x^2+6xy+y^2+x^2-2xy+y^2\)

\(=10x^2-4xy+2y^2\) 

\(7.\left(x-4\right)^2+\left(x+4y\right)\) 

\(=x^2-8x+16+x+4y\) 

\(=x^2-7x+16+4y\) 

\(10.\left(2x+7\right)^2+\left(-2x-3\right)^2\) 

\(=4x^2+28x+49+4x^2+12x+9\) 

\(=8x^2+40x+58\)

\(12.-\left(x+1\right)^2-\left(x-1\right)^2\) 

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

\(=-x^2-2x-1+x^2+2x-1\)  

\(=4x\) 

\(5.-\left(x+5\right)^2-\left(x-3\right)^2\) 

\(=-\left(x^2+10x+25\right)-\left(x^2-6x+9\right)\) 

\(=-x^2-10-25+x^2+6x-9\) 

\(=-16x-16\) 

\(8.-\left(-2x+3\right)^2-\left(5x-3\right)^2\) 

\(=4x^2+12x+9-25x^2+30x-9\) 

\(=-21x^2+42x\)

\(11.-\left(2x-y\right)^2-\left(x+3y\right)^2\) 

\(=-4x^2+4xy-y^2-\left(x^2+6xy+9y^2\right)\) 

\(=-4x^2+4xy-y^2-x^2-6xy-9y^2\) 

\(=-5x^2-2xy-10y^2\)

4: =9x^2+6xy+y^2+x^2-2xy+y^2

=10x^2+4xy+2y^2

5: =-x^2-10x-25-x^2+6x-9

=-4x-34

7; \(=x^2-8xy+16y^2+x+4y\)

10: \(=4x^2+28x+49+4x^2+12x+9\)

=8x^2+40x+58

11: =-4x^2+4xy-y^2-x^2-6xy-9y^2

=-5x^2-2xy-10y^2

1.

\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)

\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)

\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)

\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)

\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)

\(\Leftrightarrow7x^2+20x+11=0\)

2.

ĐKXĐ: ...

\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)

\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)

\(\Leftrightarrow...\)

3.

ĐKXĐ: ...

Từ pt dưới:

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)

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

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

\(\Leftrightarrow\left(x-1\right)^3=\left(y+1\right)^3\)

\(\Leftrightarrow y=x-2\)

Thế vào pt trên:

\(x^2-2x+3=2\sqrt{5x-2}+\sqrt{7x-1}\)

\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)

\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)

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

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge1\end{matrix}\right.\)

\(xy-y^2+2y-x-1=\sqrt{y-1}-\sqrt{x}\)

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

\(\Leftrightarrow\left(y-1\right)\left(x-y+1\right)+\dfrac{x-y+1}{\sqrt{x}+\sqrt{y-1}}=0\)

\(\Leftrightarrow\left(x-y+1\right)\left(y-1+\dfrac{1}{\sqrt{x}+\sqrt{y-1}}\right)=0\)

\(\Leftrightarrow x-y+1=0\)

\(\Rightarrow y=x+1\)

Thay xuống pt dưới:

\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)

\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+\left(7-x-3\sqrt{5-x}\right)=0\)

\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)

\(\Leftrightarrow...\)

ĐKXĐ: x<>7/2 và y<>-6

\(\left\{{}\begin{matrix}\dfrac{3}{2x-7}+\dfrac{4}{y+6}=7\\\dfrac{2}{2x-7}-\dfrac{3}{y+6}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{9}{2x-7}+\dfrac{12}{y+6}=21\\\dfrac{8}{2x-7}-\dfrac{12}{y+6}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{2x-7}=17\\\dfrac{2}{2x-7}-\dfrac{3}{y+6}=-1\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}x=4\\y=-5\end{matrix}\right.\left(nhận\right)\)

đk x khác 7/2 ; y khác -6 

Đặt \(\dfrac{1}{2x-7}=t;\dfrac{1}{y+6}=u\)

\(\left\{{}\begin{matrix}3t+4u=7\\2t-3u=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6t+8u=14\\6t-9u=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=1\\t=\dfrac{-1+3u}{2}=1\end{matrix}\right.\)

Theo cách đặt \(\left\{{}\begin{matrix}2x-7=1\\y+6=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-5\end{matrix}\right.\left(tm\right)\)

làm giúp mk bài này nhá                                                                                                              0+1+2+...+2017  có bao nhiêu số hạng

Giải hệ phương trình: \(\left\{\begin{matrix} xy-y^2-x+2y=\sqrt{y-1}+1-\sqrt{x} - Hy Vũ