a) Đặt \(\frac{x}{5}=\frac{y}{7}=k\)

\(\Rightarrow\hept{\begin{cases}x=5k\\y=7k\end{cases}}\)

\(\Rightarrow xy=5k.7k\)

\(\Rightarrow140=35k^2\)

\(\Rightarrow k^2=4\)

\(\Rightarrow\orbr{\begin{cases}k=2\\k=-2\end{cases}}\)

Với k = 2 ta có :

+) \(\frac{x}{5}=2\Rightarrow x=10\)

+) \(\frac{y}{7}=2\Rightarrow y=14\)

Với k = -2 ta có :

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

+) \(\frac{y}{7}=-2\Rightarrow y=-14\)

Vậy  \(\left(x;y\right)=\left\{\left(10;14\right);\left(-10;-14\right)\right\}\)

b) Ta có :

\(x:y:z\)\(=\)\(2:5:7\)\(\Rightarrow\)\(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}\)\(\Rightarrow\)\(\frac{3x}{6}=\frac{2y}{10}=\frac{z}{7}\)

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

\(\frac{3x}{6}=\frac{2y}{10}=\frac{z}{7}=\frac{3x+2y-z}{6+10-7}=\frac{27}{9}=3\)

+) \(\frac{x}{2}=3\Rightarrow x=6\)

+) \(\frac{y}{5}=3\Rightarrow y=15\)

+) \(\frac{z}{7}=3\Rightarrow z=21\)

Vậy x = 6, y = 15 và z = 21

_Chúc bạn học tốt_

a, x.y/5.7=140/35

=140/35=4

x/5=4/7

x/7=5/4

x.7=5.4

x.7=20

x=20;7

x=20/7

b,chịu

tk thì tk ko tk cx đc

a, \(\frac{x}{5}=\frac{y}{7}\left(x.y=140\right)\)

Đặt \(\frac{x}{5}=\frac{y}{7}=k\)

\(\Rightarrow7x=5y\)

\(\Rightarrow x.y=7k.5k=35k^2=140\)

\(\Rightarrow k^2=4\Rightarrow k=\pm2\)

\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x=2.7=14\\y=2.5=10\end{cases}}\\\hept{\begin{cases}x=\left(-2\right).7=-14\\y=\left(-2\right).5=-10\end{cases}}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x=2.7=14\\y=2.5=10\end{cases}}\\\hept{\begin{cases}x=\left(-2\right).7=-14\\y=\left(-2\right).5=-10\end{cases}}\end{cases}}\)

Vậy ....

b, \(x:y:z=2:5:7\left(3x+2y-z=27\right)\)

Đặt \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=k\)

\(\Leftrightarrow x=2k;y=5k=z=7k\)

\(\Leftrightarrow3x+2y-z=6k+10k-7k=27\)

\(\Leftrightarrow x=6;y=15;z=21\)

Vậy ...

Bài làm

y . 5 + y . 6 - y - 37 = 63

y ( 5 + 6 - 1 )          = 63 + 37

y . 10                      = 100

y                             = 10

ban oi 1 o dau

\(y\times5+y\times6-y-37=63\)

=> y x 5 + y x 6 - y = 63 + 37

=> y x 5 + y x 6 - y x 1 = 100

=> y x ( 5 + 6 - 1 ) = 100

=> y x 10 = 100

=> y = 100 : 10

=> y = 10

Study well ! >_<

Bài 2:

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

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

Bài 3:

=>x^2=5

hay \(x=\pm\sqrt{5}\)

\(A=x^2-6x+15\)

\(A=x^2-2\cdot x\cdot3+3^2+6\)( biến đổi về dạng HĐT )

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

vì ( x - 3 )² luôn >= 0 với mọi x

\(\Rightarrow A\ge6\)với mọi x

Dấu "=" xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy Amin = 6 <=> x = 3

\(B=2x^2-10x+8\)

\(B=2\left(x^2-5x+4\right)\)

\(B=2\left(x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{9}{4}\right)\)

\(B=2\left[\left(x-\frac{5}{2}\right)^2-\frac{9}{4}\right]\)

\(B=2\left(x-\frac{5}{2}\right)^2-\frac{9}{2}\)

Vì 2( x - 5/2 )² luôn >= 0 với mọi x

\(\Rightarrow B\ge\frac{-9}{2}\)với mọi x

Dấu "=" xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Vậy Bmin = -9/2 <=> x = 5/2

a, Có \(\dfrac{3x-2y}{7}=\dfrac{4x+3y}{5}\)

=> 5(3x-2y)=7(4x+3y)

=> 15x-10y=28x+21y

=> 15x-28x=21y+10y

=> -13x=31y

=> \(\dfrac{x}{y}=\dfrac{31}{-13}=\dfrac{-31}{13}\)

b,\(\dfrac{5x-2y}{3x+4y}=\dfrac{-3}{4}\)

=> 4(5x-2y)=-3(3x+4y)

=> 20x-8y= -9x-12y

=> 20x+9x=-12y+8y

=> 29x=-4y

=> \(\dfrac{x}{y}=\dfrac{-4}{29}\)

aVT=.\(\left(a+b+c\right)^2+a^2+b^2+c^2\)

=\(a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)

=\(2a^2+2b^2+2c^2+2ab+2ac+2bc\)

VP=\(\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2\)=\(a^2+2ab+b^2+b^2+2bc+b^2+a^2+2ac+c^2\)

=\(2a^2+2b^2+2c^2+2ab+2bc+2ac\)

Vậy VT=VP

a)\(\text{(a+b+c)^2 +a^2+b^2+c^2=(a+b)^2+(b+c)^2+(c+a)^2}\)

Ta có:

\(\left(a+b+c\right)^2+a^2+b^2+c^2=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)

\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(c^2+2ca+a^2\right)\)

\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)

Vậy \(\left(a+b+c\right)^2+a^2+b^2+c^2=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)

b) Câu b sao chỉ có một vế vậy , hằng đẳng thức thì phải có hai vế chứ

b) \(\text{x^4+y^4+(x+y)^4=2(x^2+xy+y^2)^2}\)

Ta có:

\(x^4+y^4+\left(x+y\right)^4=x^4+y^4+x^4+4x^3y+6x^2y^2+4xy^3+y^4\)

\(2x^4+2y^{\text{4}}+4x^3y+6x^2y^2+4xy^3=2\left(x^4+y^4+2x^3y+3x^2y^2+2xy^3\right)\)

\(=2\left[\left(x^2\right)^2+\left(y^2\right)^2+\left(xy\right)^2+2x^2.y^2+2y^2.xy+2x^2.xy\right]\)

\(=2\left(x^2+xy+y^2\right)^2\)

Vậy \(x^4+y^4+\left(x+y\right)^4=2\left(x^2+xy+y^2\right)^2\)