A

Chọn A

Ta có: 2x=y3=z52x=y3=z5

⇒x=y6=z25⇒x=y6=z25và x+y−z2=−20x+y−z2=−20

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

x=y6=z25=x+y−z21+6−5=−202=−10x=y6=z25=x+y−z21+6−5=−202=−10(vìx+y−z2=−20x+y−z2=−20)

⇒\hept⎧⎨⎩x=−10y=−10⋅6=−60z2=−10⋅5=−50⇒\hept⎧⎨⎩x=−10y=−60z=−100

x2=y3=z5" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:18.56px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">x2=y3=z5 và x2=y3=z5=x+y+z2+3+5=2010=2" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:18.56px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">x2=y3=z5=x+y+z2+3+5=2010=2

x2=2" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:18.56px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">x2=2 y3=2" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:18.56px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">y3=2 z5=2" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:18.56px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">z5=2 $\Rightarrow z=2.5=10$

Vậy $x=4$; $y=6$ và $z=10$.

Chúc bạn học tốt!

Ta có :

x - y + z = 20

\(\hept{\begin{cases}\frac{x}{3}=\frac{y}{2}\\\frac{y}{3}=\frac{z}{5}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{9}=\frac{y}{6}\\\frac{y}{6}=\frac{z}{10}\end{cases}}\Rightarrow\frac{x}{9}=\frac{y}{6}=\frac{z}{10}\)

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

\(\frac{x}{9}=\frac{y}{6}=\frac{z}{10}=\frac{x-y+z}{9-6+10}=\frac{20}{13}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{20}{13}.9=\frac{180}{13}\\y=\frac{20}{13}.6=\frac{120}{13}\\z=\frac{20}{13}.10=\frac{200}{13}\end{cases}}\)

a: =>xy=-18

=>x,y khác dấu

mà x<y<0 

nên không có giá trị nào của x và y thỏa mãn yêu cầu đề bài

b: =>(x+1)(y-2)=3

\(\Leftrightarrow\left(x+1,y-2\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(0;5\right);\left(2;3\right);\left(-2;-1\right);\left(-4;1\right)\right\}\)

c: \(\Leftrightarrow8x-4=3x-9\)

=>5x=-5

hay x=-1

a) \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

Thay \(x-y=7\)vào biểu thức ta được: 

\(A=7^2+2.7+37=49+14+37=100\)

b) Ta có: \(x+y=3\)\(\Rightarrow\left(x+y\right)^2=9\)\(\Rightarrow x^2+y^2+2xy=9\)

mà \(x^2+y^2=5\)\(\Rightarrow5+2xy=9\)

\(\Rightarrow2xy=4\)\(\Rightarrow xy=2\)

Vậy \(xy=2\)

a) A = x( x + 2 ) + y( y - 2 ) - 2xy + 37

= x² + 2x + y² - 2y - 2xy + 37

= ( x² - 2xy + y² ) + ( 2x - 2y ) + 37

= ( x - y )² + 2( x - y ) + 37

Thế x - y = 7 vào A ta được :

A = 7² + 2.7 + 37 = 49 + 14 + 37 = 100

Vậy A = 100 khi x - y = 7

b) x + y = 3 => ( x + y )² = 9

=> x² + 2xy + y² = 9

=> 5 + 2xy = 9 ( sử dụng gt x² + y² = 5 )

=> 2xy = 4

=> xy = 2 

a) 

A = x( x + 2 ) +y( y - 2 ) - 2xy + 37 

=\(x^2+2x+y^2-2y-2xy+37\) 

=\(x^2-2xy+y^2+2x-2y+37\)     

=\(\left(x-y\right)^2+2\left(x-y\right)+37\)   

=\(7^2+2\cdot7+37\)   

=\(49+14-37\)             

= 26 

b) 

\(x^2+y^2=5\)         

\(\left(x+y\right)^2-2xy=5\)                                               

\(3^2-2xy=5\)                                             

\(9-2xy=5\)   

\(2xy=9-5\)                       

\(2xy=4\) 

\(xy=2\)

A

Chọn A

A

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)⇒\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)

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

\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)

⇒\(\left\{{}\begin{matrix}x=3.9=27\\y=3.12=36\\z=3.20=60\end{matrix}\right.\)

a) 2+3𝑥=−15−19

3x= -15 - 19 -2

3x = -36

x= -12

b) 2𝑥−5=−17+12

2x = -17 + 12 + 5

2x = 0

x = 0

c) 10−𝑥−5=−5−7−11

-x = -5 - 7 - 11 - 10 + 5

-x = -28

x = 28

d) |𝑥|−3=0

|x|= 3

x = \(\pm\)3

e) (7−|𝑥|).(2𝑥−4)=0

th1 : ( 7 - | x| ) = 0

|x|= 7

x=\(\pm\)7

th2: ( 2x-4) = 0

2x = 4

x= 2

f) −10−(𝑥−5)+(3−𝑥)=−8

-10 - x + 5 + 3 - x = -8

-10 + 5 + 3 + 8 = 2x

2x= 6

x = 3

g) 10+3(𝑥−1)=10+6𝑥

10 + 3x - 3 = 10 + 6x

3x - 6x = 10 - 10 + 3

-3x = 3

x= -1

h) (𝑥+1)(𝑥−2)=0

th1: x+1= 0

x = -1

x-2=0

x=2

hok tốt!!!

Bài 1:

b: Tọa độ giao điểm là:

\(\left\{{}\begin{matrix}-\dfrac{1}{2}x+3=x-2\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x=5\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{10}{3}-2=\dfrac{4}{3}\end{matrix}\right.\)