\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)

\(\Leftrightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)

\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)

\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)

Mà \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\)

\(\Leftrightarrow100-x=0\Leftrightarrow x=100\)

Vậy x = 100

\(m^3+m^2-2m=0\)

\(\Leftrightarrow m^3+2m^2-m^2-2m=0\)

\(\Leftrightarrow m^2\left(m+2\right)-m\left(n+2\right)=0\)

\(\Leftrightarrow\left(m^2-m\right)\left(m+2\right)=0\)

\(\Leftrightarrow m\left(m-1\right)\left(m+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m-1=0\\m+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=0\\m=1\\m=-2\end{matrix}\right.\)

Vậy...

\(x^2\left(x-1\right)-4x^2+8x-4=0\)

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

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\)

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

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

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

Vậy x = 1 hoặc x = 2

\(5a^2-5b^2-20a+20\)

\(=5\left(a^2-4a+4\right)-5b^2\)

\(=5\left(a-2\right)^2-5b^2\)

\(=5\left[\left(a-2\right)^2-b^2\right]\)

\(=5\left[\left(a-2-b\right)\left(a-2+b\right)\right]\)

\(=5\left(a-2-b\right)\left(a-2+b\right)\)

1, \(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}+\dfrac{y}{y-x}\)

\(=\dfrac{2xy}{x^2-y^2}+\dfrac{\left(x-y\right)\left(y-x\right)+2y\left(x+y\right)}{2\left(x+y\right)\left(y-x\right)}\)

\(=\dfrac{2xy}{x^2-y^2}+\dfrac{xy-x^2-y^2+xy+2xy+2y^2}{2\left(xy-x^2+y^2-xy\right)}\)

\(=\dfrac{2xy}{x^2-y^2}+\dfrac{4xy-x^2+y^2}{2\left(y^2-x^2\right)}\)

\(=\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}+\dfrac{4xy-x^2+y^2}{2\left(y-x\right)\left(x+y\right)}\)

\(=\dfrac{-4xy}{\left(y-x\right)\left(y+x\right)}+\dfrac{4xy-x^2+y^2}{2\left(y-x\right)\left(x+y\right)}\)

\(=\dfrac{y^2-x^2}{2\left(y-x\right)\left(x+y\right)}=\dfrac{\left(y-x\right)\left(x+y\right)}{2\left(y-x\right)\left(x+y\right)}=\dfrac{1}{2}\)

2, \(x^2-y^2-2y-1\)

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

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

\(=\left(x-y-1\right)\left(x+y+1\right)\)

\(\dfrac{4a-4b}{5a+5b}.x=\dfrac{a^2-b^2}{a^2+2ab+b^2}\)

\(\Leftrightarrow\dfrac{4\left(a-b\right)}{5\left(a+b\right)}.x=\dfrac{\left(a-b\right)\left(a+b\right)}{\left(a+b\right)^2}\)

\(\Leftrightarrow x=\dfrac{a-b}{a+b}.\dfrac{5\left(a+b\right)}{4\left(a-b\right)}\)

\(\Leftrightarrow x=\dfrac{5}{4}\)

Vậy \(x=\dfrac{5}{4}\)

Bài 1:

Giải:

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

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\)

Vậy x = 6, y = 10

Bài 2:

Ta có: \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)

\(\Rightarrow ab-6a+5b-30=ab+6a-5b-30\)

\(\Rightarrow-6a+5b=6a-5b\)

\(\Rightarrow10b=12a\)

\(\Rightarrow6a=5b\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{5}{6}\)

\(\Rightarrowđpcm\)

a, Giải:

Ta có: \(\dfrac{x}{5}=\dfrac{y}{7}\Rightarrow\dfrac{7}{5}x=y\)

\(x-y=-10\)

\(\Rightarrow x-\dfrac{7}{5}x=-10\)

\(\Rightarrow\dfrac{-2}{5}x=-10\)

\(\Rightarrow x=25\Rightarrow y=35\)

Vậy x = 25, y = 35

b, Giải:

Ta có: \(3x=4y\Rightarrow\dfrac{3}{4}x=y\)

\(y+x=14\)

\(\Rightarrow\dfrac{3}{4}x+x=14\)

\(\Rightarrow\dfrac{7}{4}x=14\)

\(\Rightarrow x=8\Rightarrow y=6\)

Vậy x = 8, y = 6

c, Ta có: \(\dfrac{4}{x}=\dfrac{2}{y}\Rightarrow\dfrac{x}{4}=\dfrac{y}{2}\Rightarrow x=2y\)

\(2x-y=12\)

\(\Rightarrow4y-y=12\)

\(\Rightarrow3y=12\)

\(\Rightarrow y=4\)

\(\Rightarrow x=8\)

Vậy x = 8, y = 4

Bằng 1 vì 2 số đó không thể chia hết cho số nào ngoài 1

a, \(4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)

\(\Leftrightarrow4x^2-20x-7x^2+28x+3x^2=12\)

\(\Leftrightarrow8x=12\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy...

b, \(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)

\(\Leftrightarrow-3x^2+15x+5x-5+3x^2=4-x\)

\(\Leftrightarrow21x=9\)

\(\Leftrightarrow x=\dfrac{3}{7}\)

Vậy...

c, \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)

\(\Leftrightarrow x^2-9x+20-x^2+x+2=7\)

\(\Leftrightarrow-8x=-15\Leftrightarrow x=\dfrac{15}{8}\)

Vậy...

d, \(-\left(x+3\right)\left(x-4\right)+\left(x-1\right)\left(x+1\right)=10\)

\(\Leftrightarrow-x^2+x+12+x^2-1=10\)

\(\Leftrightarrow x=-1\)

Vậy...

e, \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)

\(\Leftrightarrow x^3-27+5x-x^3=6x\)

\(\Leftrightarrow x=-27\)

Vậy...