làm 1 bài cx đc hết càng tốt ạ

Bài 5:

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

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

\(M=\left(5x\right)^2\)

Thay \(x=-\dfrac{1}{5}\) vào biểu thức M ta có:

\(\left(5\cdot-\dfrac{1}{5}\right)^2=\left(-1\right)^2=1\)

Vậy: ...

b) \(N=\left(3x-1\right)^2-2\left(9x^2-1\right)+\left(3x+1\right)^2\)

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

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

\(N=\left(-2\right)^2=4\)

Vậy: ...

5

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

6

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

7

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

8

\(=\left(2x-1-3x+1\right)\left(2x-1+3x-1\right)=\left(-x\right)\left(5x-2\right)\)

9

\(=\left(2x\right)^2-2.2x.y+y^2=\left(2x-y\right)^2\)

10

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

11

\(=\left(x^2y^2\right)^2+2.2x^2y^2+2^2=\left(x^2y^2+2\right)\)

12

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

13

\(=1-\left(3\sqrt{3}x\right)^3=\left(1-3\sqrt{3}x\right)\left[1^2+3\sqrt{3}.x+\left(3\sqrt{3}.x\right)^2\right]=\left(1-3\sqrt{3}x\right)\left(1+3\sqrt{3}x+27x^2\right)\)

  đây

  ạ

14

\(=\left(x-3\right)^3+3^3=\left(x-3+3\right)\left[\left(x-3\right)^2-\left(x-3\right).3+3^2\right]\\ =x\left(x^2-6x+9-3x+9+9\right)=x\left(x^2-9x+27\right)\)

`A=4(3^2+1)(3^4+1)...(3^64+1)`

`=>2A=(3^2-1)(3^2+1)(3^4+1)...(3^64+1)`

- Ta có: 

`(3^2-1)(3^2+1)=3^4-1`

`(3^4-1)(3^4+1)=3^16-1`

`....`

`(3^64-1)(3^64+1)=3^128-1`

Suy ra `2A=3^128-1=B`

`=>A<B`

Bài 1:
Ta có: `2>=x+y>=2\sqrt(xy)=>sqrt(xy)<=1=>0<xy<=1`

Áp dụng BĐT Cau chy cho hai số không âm `xy` và `1/(xy)` ta có:
`A=xy+1/(xy)>=2\sqrt(xy . 1/(xy))=2`

Dấu "=" xảy ra `<=>x=y=1`

Vậy `A_(min)=2<=>x=y=1`

----

Bài 2:

Đặt `B=x(x-1)+y(y-1)`

Ta có: `B=x(x-1)+y(y-1)=(x^2-6x+9)+(y^2-6y+9)+5(x+y)-18=(x-3)^2+(y-3)^2+5(x+y)-18>=12`

Dấu "=" xảy ra `<=>x=y=3`

Vậy `B_(min)=12<=>x=y=3`

dòng cuối bài 2: có kêu tìm GTNN với có cho B đâu mà cần kết luận:v

Giải bài 2:

Ta có:

\(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)

\(\Leftrightarrow x^2+y^2-\left(x+y\right)\ge\dfrac{\left(x+y\right)^2}{2}-\left(x+y\right)\\ \Leftrightarrow x\left(x-1\right)+y\left(y-1\right)\ge\dfrac{6^2}{2}-6\\ \Leftrightarrow x\left(x-1\right)+y\left(y-1\right)\ge12\left(đpcm\right)\)

Dấu "=" xảy ra khi x = y = 3.

Bạn nên chịu gõ gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn. 

a: Xét ΔABE vuông tại E và ΔACF vuông tại F có

góc BAE chung

=>ΔABE đồng dạng với ΔACF

=>AE/AF=AB/AC

=>AE/AB=AF/AC và AE*AC=AB*AF

b: Xét ΔAEF và ΔABC có

AE/AB=AF/AC

góc FAE chung

=>ΔAEF đồng dạng với ΔABC

=>góc AFE=góc ACB

1: \(=\dfrac{13}{39}\cdot\dfrac{x^4}{x^2}=\dfrac{1}{3}x^2\)

2: \(=\dfrac{-20}{4}\cdot\dfrac{y^7}{y^4}=-5y^3\)

3: =(-z)^15-12=(-z)^3=-z^3

4: \(=\dfrac{-15}{3}\cdot\dfrac{x}{x}\cdot\dfrac{y^2}{y}\cdot\dfrac{z^3}{z^2}=-5yz\)

5: \(=\dfrac{-12}{4}\cdot\dfrac{x^5}{x^4}\cdot\dfrac{y^4}{y^2}=-3xy^2\)

6: \(=\dfrac{15}{6}\cdot\dfrac{x^2y^3z^2}{xz^2}=\dfrac{5}{2}xy^3\)

7: =(-3/5:5/6)*x^4y^4:xy

=-18/25x^3y^3

8: =-34/17*x^5y^3z^2:x^4y^3z

=-2xz

12: =(-x)^10:(-x)^7=(-x)^3=-x^3

11: =(-xy^2z)^3-2=(-xy^2z)^1=-xy^2z

`@` `\text {Ans}`

`\downarrow`

`1,`

\(13x^4\div39x^2\)

`= (13 \div 39)*(x^4 \div x^2)`

`= 1/3x^2`

`3,`

\(\left(-z\right)^{15}\div\left(-z\right)^{12}\)

`=`\(\left(-z\right)^{15-12}=\left(-z\right)^3\) 

`2,`

\(-20y^7\div4y^4\)

`=(-20 \div 4)*(y^7 \div y^4)`

`= -5y^3`

`4,`

\(15xy^2z^3\div\left(-3xyz^2\right)\)

`= [15 \div (-3)]*(x \div x)*(y^2 \div y)*(z^3 \div z^2)`

`= -5yz`

`5,`

\(12x^5y^4\div\left(-4x^4y^2\right)\)

`= [ 12 \div (-4)]*(x^5 \div x^4)*(y^4 \div y^2)`

`= -3xy^2`

`6,`

\(\left(-15xy^3z^2\right)\div\left(-6xz^2\right)\)

`= (-15 \div -6) * (x \div x) * y^3 * (z^2 \div z^2)`

`= 5/2y^3`

`7,`

\(\dfrac{3}{5}x^4y^4\div\left(-\dfrac{5}{6}xy\right)\)

`= [ 3/5 \div (-5/6)]*(x^4 \div x)*(y^4 \div y)`

`= -18/25x^3y^3`

`8,`

\(-34x^5y^3z^2\div17x^4y^3z\)

`= (-34 \div 17)*(x^5 \div x^4)*(y^3 \div y^3)*(z^2 \div z)`

`= -2xz`

`9,`

\(\left(-18xy^3z\right)^2\div\left(-36xz^2\right)\)

`= 324x^2y^6z^2 \div (-36xz^2)`

`= [ 324 \div (-36)] * (x^2 \div x)*y^6 *(z^2 \div z^2)`

`= -9xy^6`

`10,`

\(\left(-\dfrac{1}{2}a^3b^4c^5\right)\div\left(\dfrac{3}{2}a^2bc^4\right)\)

`= (-1/2 \div 3/2)*(a^3 \div a^2)*(b^4 \div b)*(c^5 \div c^4)`

`= -1/3ab^3c`

`11,`

\(\left(-xy^2z\right)^3\div\left(-xy^2z\right)^2\)

`= -x^3y^6z^3 \div (-x^2y^4z^2)`

`= y^2`

`12,`

\(x^{10}\div\left(-x\right)^7\)

`= (-x)`\(^{10-7}\)

`= (-x)^3`

`@` `\text {Kaizuu lv u.}`

\(a,5x^2y^4:\left(10x^2y\right)\\ =5x^2y^4.\dfrac{1}{10x^2y}\\ =\dfrac{5}{10}.\dfrac{x^2}{x^2}.\dfrac{y^4}{y}\\ =\dfrac{1}{2}y^3\)

\(b,\dfrac{3}{4}x^3y^3:\left(-\dfrac{1}{2}x^2y^2\right)\\ =\dfrac{3}{4}x^3y^3.\left(-\dfrac{2}{x^2y^2}\right)\\ =\dfrac{3}{4}.\left(-2\right).\dfrac{x^3}{x^2}.\dfrac{y^3}{y^2}\\ =-\dfrac{3}{2}xy\)

\(c,\left(-xy\right)^{10}:\left(-xy\right)^5=\left(-xy\right)^{10-5}=\left(-xy\right)^5=-x^5y^5\)

\(d,21xy^5z^3:\left(-7xy^2z^3\right)\\ =21xy^5z^3.\left(-\dfrac{1}{7xy^2z^3}\right)\\ =21.\left(-\dfrac{1}{7}\right).\dfrac{x}{x}.\dfrac{y^5}{y^2}.\dfrac{z^3}{z^3}\\ =-3y^3\)

a: =5/10*x^2:x^2*y^4:y=1/2y^3

b: =(3/4:(-1/2))*x^3:x^2*y^3:y^2

=-3/2xy

c: =(-xy)^10-5=(-xy)^5=-x^5y^5

d: \(=\dfrac{21}{-7}\cdot x:x\cdot\dfrac{y^5}{y^2}\cdot\dfrac{z^3}{z^3}=-3y^3\)

Giả sử x³ + y³ + z³ < 3, từ đó ta suy ra:

x³ + y³ + z³ < \(\dfrac{\left(x+y+z\right)^3}{9}\) 

Do x + y + z = 3, ta có:

x³ + y³ + z³ < \(\dfrac{27}{9}\) 

x³ + y³ + z³ < 3

\(\Rightarrow x^3+y^3+z^3\ge3\) 

\(\Rightarrowđpcm\)