bằng 16 đúng ko

= a.a+2.a+a.a+2.8

= a^2+2a+a^2+16

= 2a^2+2a+16

\(\left(a+2\right)\left(a+8\right)=a^2+8a+2a+16=a^2+10a+16\)

a) y-186=6*8

y-186=48

=> y= 48+186=234

b) 4*y:3=12

4*y=12*3

4*y=36

=> y= 36:4=9

a 234 , b 16

a) 

\(=\frac{8^3}{\left(-8\right)^{-5}}=\frac{8^3}{-\frac{1}{8^5}}=8^3.-\left(8\right)^5=-8^8\)

b)

\(=\frac{15x^2y^2}{5xy^2}=3x\)

khó thế

\(8^3:\left(-8\right)^{-5}=\frac{8^3}{-\frac{1}{8^5}}=8^3.\left(-8\right)^5=-8^8\)

\(15x^2y^2:5xy^2=3x\)

a. 8 . 2x - 5 - 32 = 119

8 . 2x - 5 - 9 = 119

8 . 2x - 5 = 119 + 9

8 . 2x - 5 = 128

2x - 5 = 128 : 8

2x -  5 = 16

2x -  5 = 24 (cùng cơ số)

x - 5 = 4

x = 4 + 5; x = 9

b. 5x + 2x = 62 - 50

7x = 36 - 1

7x = 35

x = 5

a. 8 . 2^{x - 5} - 3² = 119

8 . 2^{x - 5} - 9 = 119

8 . 2^{x - 5} = 119 + 9

8 . 2^{x - 5 =} 128

2^{x - 5} = 128 : 8

2^{x -  5} = 16

2^{x -  5} = 2⁴ (cùng cơ số)

x - 5 = 4

x = 4 + 5; x = 9

b. 5x + 2x = 6² - 5⁰

7x = 36 - 1

7x = 35

x = 6

Sai đáp án, 

b. 

x = 5

8a.

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\left(3x^2-5x+1\right)=3-5+1=-1\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(-3x+2\right)=-3+2=-1\)

\(\Rightarrow\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)\Rightarrow\) hàm có giới hạn tại \(x=1\)

Đồng thời \(\lim\limits_{x\rightarrow1}f\left(x\right)=-1\)

b.

\(\lim\limits_{x\rightarrow2^+}f\left(x\right)=\lim\limits_{x\rightarrow2^+}\dfrac{x^3-8}{x-2}=\lim\limits_{x\rightarrow2^+}\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2^+}\left(x^2+2x+4\right)=12\)

\(\lim\limits_{x\rightarrow2^-}f\left(x\right)=\lim\limits_{x\rightarrow2^-}\left(2x+1\right)=5\)

\(\Rightarrow\lim\limits_{x\rightarrow2^+}f\left(x\right)\ne\lim\limits_{x\rightarrow2^-}f\left(x\right)\Rightarrow\) hàm ko có giới hạn tại x=2

9.

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{x^2+mx+2m+1}{x+1}=\dfrac{0+0+2m+1}{0+1}=2m+1\)

\(\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2x+3m-1}{\sqrt{1-x}+2}=\dfrac{0+3m-1}{1+2}=\dfrac{3m-1}{3}\)

Hàm có giới hạn khi \(x\rightarrow0\) khi:

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)\Rightarrow2m+1=\dfrac{3m-1}{3}\)

\(\Rightarrow m=-\dfrac{4}{3}\)

a) 4/5 dm3= 800 cm3

B) 5/8 dm3= 625 cm3

c)3/4 dm3= 750 cm3

d)1/8 dm3= 125 cm3

B) 5/8 dm3= 625 cm3

c)3/4 dm3= 750 cm3

d)1/8 dm3= 125 cm3

k mik nha bn

áp dụng đl ta-lét vào tam giác có:

\(\dfrac{BC}{CA}=\dfrac{DE}{EA}=\dfrac{BC}{5}=\dfrac{3}{8}=>BC=\dfrac{3}{8}.5=\dfrac{15}{8}=1,875\)

X = BC + CA = 1,875 + 5 = 6,875

`x-1/9 =8/3`

`=>x=8/3 +1/9`

`=> x= 24/9 +1/9`

`=>x= 25/9`

Vậy `x=25/9`

__

`x-2/20=-5/2-x`

`=>x+x=-5/2 +2/20`

`=> 2x= -50/20 +2/20`

`=> 2x= -48/20`

`=> x= -12/5:2`

`=>x=-12/5 xx1/2`

`=>x= -12/10`

`=>x= -6/5`

Vậy `x=-6/5`

`@` `\text {Ans}`

`\downarrow`

`a)`

Đề là \(\dfrac{x-1}{9}=\dfrac{8}{3}\) phải hongg bạn?

\(\dfrac{x-1}{9}=\dfrac{8}{3}\)

`=>` `(x-1)3 = 8*9`

`=> (x-1)3=72`

`=> x-1=72 \div 3`

`=> x-1=24`

`=> x=25`

`b)`

\(\dfrac{x-2}{20}=\dfrac{-5}{2-x}\)

`=>` `(x-2)(2-x)=20*(-5)`

`=> (x-2)(2-x)=-100`

`=> -[(x-2)(x-2)]=-100`

`=> -(x-2)^2 = -100`

`=> (x-2)^2 = 100`

`=> (x-2)^2 = (+-10)^2`

`=>`\(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)