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$\color{#0000CD}{\text{A}}$   $\color{#0000FF}{\text{n}}$
$\color{#8A2BE2}{\text{nn}}$

a: ĐKXĐ: \(x\ne0;y\ne0\)

Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\)

Hệ phương trình sẽ trở thành: \(\left\{{}\begin{matrix}a-2b=-1\\2a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-2b=-1\\4a+2b=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a-2b+4a+2b=-1+6\\2a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a=5\\b=3-2a\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=1\\b=3-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=1\\\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\left(nhận\right)\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne y\\x\ne-\dfrac{y}{2}\end{matrix}\right.\)

Đặt \(\dfrac{1}{x-y}=a;\dfrac{1}{2x+y}=b\)

Hệ phương trình sẽ trở thành:

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

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

=>\(\left\{{}\begin{matrix}a=\dfrac{2}{5}\\b=2-\dfrac{2}{5}=\dfrac{10}{5}-\dfrac{2}{5}=\dfrac{8}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x-y}=\dfrac{2}{5}\\\dfrac{1}{2x+y}=\dfrac{8}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{5}{2}\\2x+y=\dfrac{5}{8}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-y+2x+y=\dfrac{5}{2}+\dfrac{5}{8}\\x-y=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{25}{8}\\y=x-\dfrac{5}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{25}{8}:3=\dfrac{25}{24}\\y=\dfrac{25}{24}-\dfrac{5}{2}=\dfrac{25}{24}-\dfrac{60}{24}=-\dfrac{35}{24}\end{matrix}\right.\left(nhận\right)\)

c: ĐKXĐ: \(x\ne1;y\ne-3\)

Đặt \(\dfrac{x}{x-1}=a;\dfrac{1}{y+3}=b\)

Hệ phương trình sẽ trở thành:

\(\left\{{}\begin{matrix}3a-2b=3\\4a+b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-2b=3\\8a+2b=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3a-2b+8a+2b=13\\4a+b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11a=13\\b=5-4a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{13}{11}\\b=5-4\cdot\dfrac{13}{11}=\dfrac{55}{11}-\dfrac{52}{11}=\dfrac{3}{11}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{x-1}=\dfrac{13}{11}\\\dfrac{1}{y+3}=\dfrac{3}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13\left(x-1\right)=11x\\y+3=\dfrac{11}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x-13=11x\\y=\dfrac{11}{3}-3=\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{13}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\left(nhận\right)\)

a) 

\(\left(-12,5\right)+3.4+12,5+\left(-3,4\right)\\ =\left(12,5-12,5\right)+\left(3,4-3,4\right)\\ =0+0=0\)

b) 

\(32,8+4,2+\left(-4,3\right)+\left(-32,8\right)+4,3\\ =\left(32,8-32,8\right)+\left(4,3-4,3\right)+4,2\\ =0+0+4,2\\ =4,2\)

c) 

\(-\left(42,5+150\right)\cdot2,5-7,5\cdot2,5\\ =2,5\left(-42,5-150-7,5\right)\\ =2,5\cdot\left(-50-150\right)\\ =2,5\cdot-200\\ =-500\) 

d) 

\(\left(-2,45\right)\cdot2,6+2,6\cdot\left(-7,55\right)\\ =2,6\cdot\left(-2,45-7,55\right)\\ =2,6\cdot-10\\ =-26\)

a: \(\left(-12,5\right)+3,4+12,5+\left(-3,4\right)\)

\(=\left(-12,5+12,5\right)+\left(3,4-3,4\right)\)

=0+0=0

b: \(32,8+4,2+\left(-4,3\right)+\left(-32,8\right)+4,3\)

\(=\left(32,8-32,8\right)+\left(4,3-4,3\right)+4,2\)

=0+0+4,2

=4,2

c: \(-\left(42,5+150\right)\cdot2,5-7,5\cdot2,5\)

\(=2,5\cdot\left(-42,5-150-7,5\right)\)

\(=2,5\cdot\left(-200\right)=-500\)

d: Sửa đề: \(\left(-2,45\right)\cdot2,6+2,6\cdot\left(-7,55\right)\)

\(=2,6\left(-2,45-7,55\right)\)

\(=2,6\cdot\left(-10\right)=-26\)

2:

a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)

b: \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=\left(x+5\right)^2\)

c: \(x^2+12x+36=x^2+2\cdot x\cdot6+6^2=\left(x+6\right)^2\)

d: \(4x^2+4x+1=\left(2x\right)^2+2\cdot2x\cdot1+1^2=\left(2x+1\right)^2\)

e: \(9x^2+6x+1=\left(3x\right)^2+2\cdot3x\cdot1+1^2=\left(3x+1\right)^2\)

f: \(16x^2+24x+9=\left(4x\right)^2+2\cdot4x\cdot3+3^2=\left(4x+3\right)^2\)

3:

a: \(A=\left(x+2\right)^2-x\left(x+3\right)+4x-3\)

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

=5x+1

b: \(B=\left(x+3\right)^2-x\left(x-5\right)+7x-8\)

\(=x^2+6x+9-x^2+5x+7x-8\)

=18x+1

c: \(C=\left(2x+3\right)^2-x\left(x+4\right)-9x-3\)

\(=4x^2+12x+9-x^2-4x-9x-3\)

\(=3x^2-x+6\)

d: \(D=\left(2x+21\right)^2-2x\left(2x-4\right)-5x-21\)

\(=4x^2+84x+441-4x^2+8x-5x-21\)

=87x+420

2:

\(a.x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\\ b.x^2+10x+25=x^2+2\cdot x\cdot5+5^2=\left(x+5\right)^2\\ c.x^2+12x+36=x^2+2\cdot x\cdot6+6^2=\left(x+6\right)^2\\ d.4x^2+4x+1=\left(2x\right)^2+2\cdot2x\cdot1+1^2=\left(2x+1\right)^2\\ e.9x^2+6x+1=\left(3x\right)^2+2\cdot3x\cdot1+1^2=\left(3x+1\right)^2\\ f.16x^2+24x+9=\left(4x\right)^2+2\cdot4x\cdot3+3^2=\left(4x+3\right)^2\)

Bài 3:

a: \(\left\{{}\begin{matrix}\dfrac{x+y}{2}=\dfrac{x-y}{4}\\\dfrac{x}{3}=\dfrac{y}{5}+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)=x-y\\\dfrac{5x}{15}=\dfrac{3y}{15}+\dfrac{15}{15}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+2y=x-y\\5x=3y+15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=0\\5x-3y=15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+3y+5x-3y=0+15\\x=-3y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}6x=15\\x=-3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{15}{6}=\dfrac{5}{2}\\y=\dfrac{x}{-3}=\dfrac{5}{2}:\left(-3\right)=-\dfrac{5}{6}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\left(x-1\right)\left(y+3\right)=xy+27\\\left(x-2\right)\left(y+1\right)=xy+8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy+3x-y-3=xy+27\\xy+x-2y-2=xy+8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-y=30\\x-2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-2y=60\\x-2y=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}6x-2y-x+2y=60-10\\x-2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=50\\2y=x-10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=10\\2y=10-10=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=0\end{matrix}\right.\)

Giúp mấy bài còn lại giùm mình với ạ

a: \(\left(2x-1\right)^4=81\)

=>\(\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b:Sửa đề: \(\left(x-1\right)^5=-32\)

=>\(\left(x-1\right)^5=\left(-2\right)^5\)

=>x-1=-2

=>x=-1

c: \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

=>\(\left(2x-1\right)^8-\left(2x-1\right)^6=0\)

=>\(\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)

=>\(\left(2x-1\right)^6\cdot\left(2x-1-1\right)\cdot\left(2x-1+1\right)=0\)

=>\(2x\left(2x-1\right)^6\cdot\left(2x-2\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

a) 

\(\left(2x-1\right)^4=81\\ \Rightarrow\left(2x-1\right)^4=3^4\)

TH1: 2x - 1 = 3 => 2x = 4 => x = 2 

TH2: 2x - 1 = -3 => 2x = -3 + 1 = -2 => x = -1 

b)

\(\left(x-1\right)^5=-32\\ \Rightarrow\left(x-1\right)^5=\left(-2\right)^5\\ \Rightarrow x-1=-2\\ \Rightarrow x=-2+1\\ \Rightarrow x=-1\)

c) 

\(\left(2x-1\right)^6=\left(2x-1\right)^8\\ \Rightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\\\Rightarrow \left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)

TH1: 

\(\left(2x-1\right)^6=0\\ \Rightarrow2x-1=0\\ \Rightarrow2x=1\\ \Rightarrow x=\dfrac{1}{2}\)

TH2: 

\(\left(2x-1\right)^2-1=0\\ \Rightarrow\left(2x-1\right)^2=1\\ \Rightarrow\left(2x-1\right)^2=1^2\)

+) 2x - 1 = 1 => 2x = 2 => x = 1

+) 2x - 1 = -1 => 2x = 0 => x = 0

a: \(\left(-\dfrac{2}{3}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{2}{3}-\dfrac{3}{7}-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left(-1+\dfrac{1}{7}\right)\cdot\dfrac{5}{4}=\dfrac{-6}{7}\cdot\dfrac{5}{4}=\dfrac{-30}{28}=-\dfrac{15}{14}\)

b: \(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)

\(=\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)\)

\(=\dfrac{5}{9}\cdot\dfrac{-22}{3}+\dfrac{5}{9}:\dfrac{-9}{15}\)

\(=\dfrac{5}{9}\cdot\dfrac{-22}{3}+\dfrac{5}{9}\cdot\dfrac{15}{-9}\)

\(=\dfrac{5}{9}\left(-\dfrac{22}{3}-\dfrac{5}{3}\right)=\dfrac{5}{9}\cdot\dfrac{-27}{3}=\dfrac{5}{9}\cdot\left(-9\right)=-5\)

c: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\left(\dfrac{12}{12}+\dfrac{8}{12}-\dfrac{3}{12}\right)\cdot\left(\dfrac{16}{20}-\dfrac{15}{20}\right)^2\)

\(=\dfrac{17}{12}\cdot\left(\dfrac{1}{20}\right)^2=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

d: \(2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^2=2:\left(\dfrac{3}{6}-\dfrac{4}{6}\right)^2\)

\(=2:\left(-\dfrac{1}{6}\right)^2=2:\dfrac{1}{36}=72\)

a; (- \(\dfrac{2}{3}\) - \(\dfrac{3}{7}\)): \(\dfrac{4}{5}\) + (- \(\dfrac{1}{3}\) + \(\dfrac{4}{7}\)): \(\dfrac{4}{5}\)

= (- \(\dfrac{2}{3}\) - \(\dfrac{3}{7}\)) x \(\dfrac{5}{4}\) + (- \(\dfrac{1}{3}\) + \(\dfrac{4}{7}\)) x \(\dfrac{5}{4}\)

= (- \(\dfrac{2}{3}\) - \(\dfrac{3}{7}\) - \(\dfrac{1}{3}\) + \(\dfrac{4}{7}\)) x \(\dfrac{5}{4}\)

= [ (- \(\dfrac{2}{3}\) - \(\dfrac{1}{3}\)) - (\(\dfrac{3}{7}\) - \(\dfrac{4}{7}\))] x \(\dfrac{5}{4}\)

= [ - 1 + \(\dfrac{1}{7}\)] x \(\dfrac{5}{4}\)

= [- \(\dfrac{7}{7}\) + \(\dfrac{1}{7}\)] x \(\dfrac{5}{4}\)

= - \(\dfrac{6}{7}\) x \(\dfrac{5}{4}\)

= - \(\dfrac{15}{14}\)

0, 12 = \(\dfrac{12}{100}\)

0, 08 = \(\dfrac{8}{100}\)

71, 238 = \(\dfrac{71238}{1000}\)

Giải thích:

Bạn viết hết cả số thập phân ở phần tử số và sau đó đếm xem ở số thập phân có bao nhiêu số thì viết 1 chữ số 1 và còn lại là số 0, ví dụ như 0, 8 = ?

Bạn sẽ lấy \(8\) đặt lên tử số \(\dfrac{8}{?}\) và ở mẫu số thì bạn đếm 0, 8 có 2 số, tương ứng với: 1 chữ số 1 và 1 chữ số 0.

\(#FallenAngel\)

\(0,12=\dfrac{12}{100}\)

\(0,08=\dfrac{8}{100}\)

\(71,238=\dfrac{71238}{1000}\)

a: \(\left\{{}\begin{matrix}x-3y=4\\2x+3y=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3y+2x+3y=4+17\\x-3y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=21\\3y=x-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\3y=7-4=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}2x+5y=8\\2x-7y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+5y-2x+7y=8-0\\2x=7y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12y=8\\x=\dfrac{7}{2}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{8}{12}=\dfrac{2}{3}\\x=\dfrac{7}{2}\cdot\dfrac{2}{3}=\dfrac{7}{3}\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}3x+5y=14\\-4x+3y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x+20y=56\\-12x+9y=60\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x+20y-12x+9y=56+60\\3x+5y=14\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}29y=116\\3x=14-5y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\3x=14-5\cdot4=14-20=-6\end{matrix}\right.\)

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

Đổi đơn vị: 27km/h= \(\dfrac{27000}{3600}\)km/s = 7,5 km/s

1 phút 15 giây = 75 giây

Quãng đường tàu hỏa đi được: s = vt=7,5 x 75 = 562,5 m

Chiều dài toa tàu: L = 562.5 - 85 = 477,5 m

đổi lại lời giải cuối: chiều dài cây cầu