x : y : z = 3 : 5 :( -2 )
\(\Rightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\)

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

\(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}=\frac{5x-y+3z}{15-5+\left(-6\right)}=\frac{-16}{4}=-4\)

\(\Rightarrow x=-12;y=-20;z=8\)

a. -11/11

   -111/1

b. -1/111

c. -1111

Chúc bạn học tốt 😽

1/2 x⁷.y¹¹.z²

a, <=> x^8 = x^7

<=> x = x^7 : x^7 (vì x khác 0)

<=> x=1

b, Nếu x=0 thì tm 

Nếu x khác 0 thì <=> 25= x^10 : x^8 <=> 25 = x^2 <=> x= 5 hoặc x= -5

c, <=> 2^x ( 1+2^3) = 144

<=> 2^x . 9 =144 <=> 2^x = 16

<=> x=4

d, Nếu x=0 thì tm

Nếu x khác 0 thì <=> 5/7 = x^2 : x <=> x=5/7

(x⁴)² = \(\frac{x^{12}}{x^5}\)

x^4.2 = x¹²:x⁵

x⁸ = x^12-5

x⁸ = x⁷

suy ra x⁸ - x⁷ = 0

        x⁷ . x - x⁷ = 0

        x⁷ . ( x -1 ) = 0

suy ra x⁷=0 hoặc x-1 =0

Xét x⁷=0 suy ra x⁷=0⁷suy ra x = 0

Xét x -1 =0 thì x = 0 + 1 = 1

Vậy x = 0 hoặc x=1

c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\cdot\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

\(\Leftrightarrow x+1=0\cdot\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{13}\right)\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

d) \(\frac{x+4}{2017}+\frac{x+3}{2018}=\frac{x+2}{2019}+\frac{x+1}{2020}\)

\(\Leftrightarrow\left(\frac{x+4}{2017}+1\right)+\left(\frac{x+3}{2018}+1\right)=\left(\frac{x+2}{2019}+1\right)+\left(\frac{x+1}{2020}\right)\)

\(\Leftrightarrow\frac{x+2021}{2017}+\frac{x+2021}{2018}-\frac{x+2021}{2019}-\frac{x+2021}{2020}=0\)

\(\Leftrightarrow\left(x+2021\right)\cdot\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)

\(\Leftrightarrow x+2021=0\)

\(\Leftrightarrow x=-2021\)

a) 2/3.X+4=-12

2/3.X=-12-4

2/3.X=-16

x=-16:2/3

x=-16.3/2

x=-24

vậy x=-24

B) 3/4+1/4:X=-3

1/4:X=-3-3/4

1/4:X=-15/4

X=1/4:-15/4

X=1/4.4/-15

X=-1/15

VẬY X=-1/15

MÌNH MỚI NGHĨ RA HAI Ý THÔI.

THÔNG CẢM CHO MÌNH BẠN NHÁ

Có |x-1| >=0 <=> |x-1| + 3 >= 3

<=>A=  \(\frac{1}{\left|x-1\right|+3}\)<=\(\frac{1}{3}\) 

Dấu "=" xảy ra <=> |x-1| = 0 <=> x=1

Vậy Max A = 1/3 <=> x=1

Xét ở mẫu , ta thấy :

\(\left|x-1\right|\ge0\)

\(\Leftrightarrow\left|x-1\right|+3\ge3\)

\(\Rightarrow\frac{1}{\left|x-1\right|+3}\le\frac{1}{3}\)

\(\Rightarrow Max=\frac{1}{3}\)

\(\Leftrightarrow x-1=0\)

<=> x = 1 

3^{n + 3} + 3^{n + 1} + 2^{n + 3} + 2^{n + 2}

= 3ⁿ.3³ + 3ⁿ.3 + 2ⁿ.2³ + 2ⁿ.2²

= 3ⁿ.(27 + 3) + 2ⁿ.(8 + 4)

= 3ⁿ.30 + 2ⁿ.12

= 3ⁿ.5.6 + 2ⁿ.2.6

= 6.(3ⁿ.5 + 2ⁿ.2)  \(⋮\)  6

Cảm ơn bạn kayasari nhiều nha !

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)

\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{101}\)

\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{99}{202}< \frac{3}{4}\)

\(\Rightarrow A< \frac{3}{4}\left(đpcm\right)\)

A= 1/4 +1/3^2 +1/4^2 +.....+ 1/100^2

< 1/4 + 1/2.3 + 1/3.4 +.....+1/99.100

=1/4 + 1/2-1/3+1/3-1/4+......+1/99-1/100

=1/4 +1/2 - 1/100 < 1/4+1/2 = 3/4

=> ĐPCM

x² + x = 0

x ( x + 1 ) = 0

=> x = 0

hoặc x + 1 = 0

x = 0 - 1

x = -1

Vậy x = 0 hoặc x = -1

\(x^2+x=0\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

vậy x =0 : x=-1