Toán

1: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=b\cdot k;c=d\cdot k\)

\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)

Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)

2: \(\dfrac{2a+b}{a-2b}=\dfrac{2\cdot bk+b}{bk-2b}=\dfrac{b\left(2k+1\right)}{b\left(k-2\right)}=\dfrac{2k+1}{k-2}\)

\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{d\left(2k+1\right)}{d\left(k-2\right)}=\dfrac{2k+1}{k-2}\)

Do đó: \(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)

3: \(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\cdot\left(k-1\right)}=\dfrac{k+1}{k-1}\)

\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)

Do đó: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

4: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5\cdot bk+3b}{5dk+3d}=\dfrac{b\left(5k+3\right)}{d\left(5k+3\right)}=\dfrac{b}{d}\)

\(\dfrac{5a-3b}{5c-3d}=\dfrac{5\cdot bk-3b}{5\cdot dk-3d}=\dfrac{b\left(5k-3\right)}{d\left(5k-3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}9x=\sqrt{2}+7\\y=\dfrac{1-\left(\sqrt{2}-1\right)x}{2}\end{matrix}\right.\)

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

ĐKXĐ: \(x\ne4;y\ne-1\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x-4}=u\\\dfrac{1}{y+1}=u\end{matrix}\right.\) ta được:

\(\left\{{}\begin{matrix}3u+2v=\dfrac{15}{12}\\2u-v=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3u+2v=\dfrac{15}{12}\\4u-2v=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7u=-\dfrac{11}{4}\\v=2u+2\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}u=-\dfrac{11}{28}\\v=\dfrac{17}{14}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-4}=-\dfrac{11}{28}\\\dfrac{1}{y+1}=\dfrac{17}{14}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-4=-\dfrac{28}{11}\\y+1=\dfrac{14}{17}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{16}{11}\\y=-\dfrac{3}{17}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{3}{x-4}+\dfrac{2}{y+1}=\dfrac{15}{12}\\\dfrac{2}{x-4}-\dfrac{1}{y+1}=-2\end{matrix}\right.\)

Đặt: \(\left\{{}\begin{matrix}a=\dfrac{1}{x-4}\\b=\dfrac{1}{y+1}\end{matrix}\right.\)

Hệ phương trình: \(\Leftrightarrow\left\{{}\begin{matrix}3a+2b=\dfrac{15}{12}\\2a-b=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a+2b=\dfrac{15}{12}\\4a-2b=-4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}7a=-\dfrac{11}{4}\\2a-b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{11}{28}\\2\cdot\left(-\dfrac{11}{28}\right)-b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{11}{28}\\-\dfrac{11}{14}-b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{11}{28}\\b=\dfrac{17}{14}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-4}=-\dfrac{11}{28}\\\dfrac{1}{y+1}=\dfrac{17}{14}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-4=\dfrac{1}{-\dfrac{11}{28}}\\y+1=\dfrac{1}{\dfrac{17}{14}}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{16}{11}\\y=-\dfrac{3}{17}\end{matrix}\right..}\)

ĐKXĐ: \(x\ne1;y\ne1\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x-1}=u\\\dfrac{1}{y-1}=v\end{matrix}\right.\) hệ trở thành:

\(\left\{{}\begin{matrix}5u+v=10\\u-3v=18\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}15u+3v=30\\u-3v=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}16u=48\\v=\dfrac{u-18}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=3\\v=-5\end{matrix}\right.\)

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

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)

Do B là giao điểm (d) với Ox

\(\Rightarrow y_B=0\Rightarrow\left(m+1\right)x_B+3=0\Rightarrow x_B=-\dfrac{3}{m+1}\) (với \(m\ne-1\))

\(\Rightarrow OB=\left|x_B\right|=\dfrac{3}{\left|m+1\right|}\)

Pt hoành độ giao điểm (d) và (d'):

\(\left(m+1\right)x+3=2x+3\Rightarrow x=0\)

\(\Rightarrow y_A=2.0+3=3\) \(\Rightarrow OA=\left|y_A\right|=3\)

\(OA=2OB\Rightarrow3=\dfrac{6}{\left|m+1\right|}\Rightarrow\left|m+1\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}m+1=2\\m+1=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=1\\m=-3\end{matrix}\right.\)

\(\dfrac{a}{b}=\dfrac{c}{d}\)

=>\(\dfrac{b}{a}=\dfrac{d}{c}\)

=>\(\dfrac{b}{a}+1=\dfrac{d}{c}+1\)

=>\(\dfrac{b+a}{a}=\dfrac{d+c}{c}\)

=>\(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Rightarrow\dfrac{b}{a}=\dfrac{d}{c}\)

\(\Rightarrow\dfrac{b}{a}+1=\dfrac{d}{c}+1\)

\(\Rightarrow\dfrac{b+a}{a}=\dfrac{d+c}{c}\)

\(\Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\left(đpcm\right)\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{2}=\dfrac{b}{4}=\dfrac{c}{6}=\dfrac{b+c-a}{4+6-2}=\dfrac{8}{8}=1\)

=>\(a=2\cdot1=2;b=1\cdot4=4;c=6\cdot1=6\)

Vậy: Số cần tìm là 246

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{2}=\dfrac{b}{4}=\dfrac{c}{6}=\dfrac{b+c-a}{4+6-2}=\dfrac{8}{8}=1\)

\(\Rightarrow a=2\cdot1=2\)

\(\Rightarrow b=4\cdot1=4\)

\(\Rightarrow c=6\cdot1=6\)

Vậy \(\left(a;b;c\right)=\left(2;4;6\right)\)

a: \(x^2-\left(m-2\right)x+m-5=0\)

\(\text{Δ}=\left(-m+2\right)^2-4\left(m-5\right)\)

\(=m^2-4m+4-4m+20\)

\(=m^2-8m+24\)

\(=m^2-8m+16+8=\left(m-4\right)^2+8>=8>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

b: Để phương trình có hai nghiệm trái dấu thì a*c<0

=>\(1\cdot\left(m-5\right)< 0\)

=>m-5<0

=>m<5

2x=3y

=>\(\dfrac{x}{3}=\dfrac{y}{2}\)

=>\(\dfrac{x}{6}=\dfrac{y}{4}\)

\(5y=4z\)

=>\(\dfrac{y}{4}=\dfrac{z}{5}\)

=>\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{5}\)

mà x+y+z=-30

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y+z}{6+4+5}=\dfrac{-30}{15}=-2\)

=>\(x=-2\cdot6=-12;y=-2\cdot4=-8;z=-2\cdot5=-10\)

Bài 2:

\(x^2+\left(m+2\right)x+2m=0\)

\(\text{Δ}=\left(m+2\right)^2-4\cdot1\cdot2m\)

\(=m^2+4m+4-8m=m^2-4m+4\)

\(=\left(m-2\right)^2>=0\forall m\)

=>Phương trình luôn có hai nghiệm x1;x2

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(m+2\right)}{1}=-m-2\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{2m}{1}=2m\end{matrix}\right.\)

\(2\cdot\left(x_1+x_2\right)+x_1x_2\)

\(=2\left(-m-2\right)+2m\)

=-2m-4+2m

=-4

=>Đây là hệ thức cần tìm

Bài 3:

a: Thay x=-2 vào phương trình, ta được:

\(\left(2m-1\right)\cdot\left(-2\right)^2+\left(m-3\right)\cdot\left(-2\right)-6m-2=0\)

=>\(4\left(2m-1\right)-2\left(m-3\right)-6m-2=0\)

=>8m-4-2m+6-6m-2=0

=>0=0

=>Phương trình luôn có nghiệm x=-2

b: TH1: m=1/2

Phương trình lúc này sẽ là:

\(\left(2\cdot\dfrac{1}{2}-1\right)\cdot x^2+\left(\dfrac{1}{2}-3\right)x-6\cdot\dfrac{1}{2}-2=0\)

\(\Leftrightarrow-\dfrac{5}{2}x-5=0\)

=>\(-\dfrac{5}{2}x=5\)

=>\(x=-5:\dfrac{5}{2}=-2\)

TH2: m<>1/2

\(\text{Δ}=\left(m-3\right)^2-4\left(2m-1\right)\left(-6m-2\right)\)

\(=m^2-6m+9+4\left(2m-1\right)\left(6m+2\right)\)

\(=m^2-6m+9+4\left(12m^2+4m-6m-2\right)\)

\(=m^2-6m+9+4\left(12m^2-2m-2\right)\)

\(=m^2-6m+9+48m^2-8m-8\)

\(=49m^2-14m+1=\left(7m-1\right)^2>=0\forall m\)

=>Phương trình luôn có hai nghiệm là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-\left(m-3\right)-\sqrt{\left(7m-1\right)^2}}{2\cdot\left(2m-1\right)}=\dfrac{-\left(m-3\right)-\left|7m-1\right|}{4m-2}\\x_2=\dfrac{-\left(m-3\right)+\sqrt{\left(7m-1\right)^2}}{2\left(2m-1\right)}=\dfrac{-\left(m-3\right)+\left|7m-1\right|}{4m-2}\end{matrix}\right.\)