\(1,\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=15\end{matrix}\right.\\ 2,7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-12\\y=-28\end{matrix}\right.\\ 3,\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=-\dfrac{9}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{45}{2}\\y=-27\\z=-\dfrac{63}{2}\end{matrix}\right.\\ 4,x:y:z=3:5:7\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)

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

\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=\dfrac{-9}{2}\)

\(x=\dfrac{-45}{2}\)

\(y=-27\)

\(z=\dfrac{-63}{2}\)

(2x+1)(3y-2)=-11.5

                     =11.-5

2x+1=5

=> x=2

3y-2=(-11)

=> y=(-3)

Con n truong hop khac thi cau tu trinh bay nhung ko ra x,y la Z

Nho tick nha

Bài 4:

a: \(=7xy\left(2-3-4\right)=-35xy\)

b: \(=\left(x-5\right)\left(x+y\right)\)

c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)

d: \(=\left(x+y\right)^3-\left(x+y\right)\)

=(x+y)(x+y+1)(x+y-1)

e: =x^2+8x-x-8

=(x+8)(x-1)

f: \(=2x^2-4x+x-2=\left(x-2\right)\left(2x+1\right)\)

g: =-5x^2+15x+x-3

=(x-3)(-5x+1)

h: =x^2-3xy+xy-3y^2

=x(x-3y)+y(x-3y)

=(x-3y)*(x+y)

Bài 4:

a: \(=7xy\left(2-3-4\right)=-35xy\)

b: \(=\left(x-5\right)\left(x+y\right)\)

c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)

d: \(=\left(x+y\right)^3-\left(x+y\right)\)

=(x+y)(x+y+1)(x+y-1)

e: =x^2+8x-x-8

=(x+8)(x-1)

f: \(=2x^2-4x+x-2=\left(x-2\right)\left(2x+1\right)\)

g: =-5x^2+15x+x-3

=(x-3)(-5x+1)

h: =x^2-3xy+xy-3y^2

=x(x-3y)+y(x-3y)

=(x-3y)*(x+y)

1

2x . 3=3y .4

=> x=2y=>\(\frac{x}{2}=y\Rightarrow\frac{x}{4}=\frac{y}{2}\)

\(\frac{x}{4}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{2}=\frac{z}{5}=\frac{x-2y+3z}{4-4+15}=\frac{1}{15}=\)

x=1/15x4=4/15

y=1/15x2=2/15

z=1/15x6=1/10

\(\Rightarrow x-y-z=\frac{4}{15}-\frac{2}{15}-\frac{1}{10}=\frac{1}{30}\)

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

4\(x^2\)-12x+9-2(9\(x^2\)+6x+1)=2\(x^2\)-4x+\(x^2\)+2x-x-2

4\(x^2\)-12x+9-18\(x^2\)-12x-2=2\(x^2\)-4x+\(x^2\)+2x-x-2

(4\(x^2\)-18\(x^2\)-2\(x^2\)-\(x^2\)) +(-12x-12x+4x-2x+x)+(9-2+2)=0

-17\(x^2\)-21x+9=0

mấy bạn giỏi toán ơi giúp mk vs

Nguyễn Thanh Hằng Nhã Doanh ngonhuminh nguyen thi vang mấy ban giup mk voi

a/Vì \(\sqrt{x^6+1}=x^2\Rightarrow\left(\sqrt{1+x^6}\right)^2=x^4\Rightarrow1+x^6=x^4\)

\(\Rightarrow1=x^4-x^6\)

\(\Rightarrow1=x^4-x^6\)(Vô lí)

Vì \(1>0\) và \(x^4-x^6< 0\) (với mọi x)

Vậy ko tìm đc giá trị x thỏa mãn

b/\(\left(2x+1\right)^5=\left(2x+1\right)^{2010}\)

\(\Rightarrow\left(2x+1\right)^{2005+5}-\left(2x+1\right)^5=0\)

\(\Rightarrow\left(2x+1\right)^5\cdot\left[\left(2x+1\right)^{2005}-1\right]=0\)

\(\Rightarrow\left(2x+1\right)^5\cdot\left(2x+1-1\right)\cdot\left(2x+1+1\right)=0\\\)

\(\Rightarrow\left\{{}\begin{matrix}\left(2x+1\right)^5=0\\2x+1-1=0\\2x+1+1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x+1=0\Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\\2x=0\Rightarrow x=0\\2x+2=0\Rightarrow2x=-2\Rightarrow x=-1\end{matrix}\right.\)

Vậy...

c/Nếu \(x-3\ge0\) hay \(x\ge3\) ta sẽ có : \(\left|x-3\right|=x-3\)

Khi đó phương trình ( hay đề bài,đẳng thức) có dạng:

\(12-\left(x-3\right)=5x+8\)

\(\Rightarrow12-x+3=5x+8\)

\(\Rightarrow15-8=5x+x\)

\(\Rightarrow7=6x\Rightarrow x=\dfrac{7}{6}\) (không thỏa mãn \(x\ge3\))

Nếu \(x-3< 0\) hay \(x< 3\) ta sẽ có: \(\left|x-3\right|=3-x\)

Khi đó phương trình có dạng:

\(12-\left(3-x\right)=5x+8\)

\(\Rightarrow9-8=5x-x\)

\(\Rightarrow1=4x\Rightarrow x=\dfrac{1}{4}\) (thỏa mãn)

Vậy....

khó + lười + nhiều = không làm

Hello

ko thích làm

a) 3y +xy+2x+6=0

    3.(y + 2) + x.(y + 2) = 0

   (3 + x).(y + 2) = 0

\(\Rightarrow\hept{\begin{cases}3+x=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-3\\y=-2\end{cases}}}\)

Vậy...