`@` `\text {Ans}`

`\downarrow`

`a,`

`x - 280 \div 35 = 554`

`=> x - 8 = 554`

`=> x = 554 + 8`

`=> x = 562`

Vậy, `x = 562`

`b,`

`x + (5 \div 3) = 500`

`=> x + 5/3 = 500`

`=> x = 500 - 5/3`

`=> x = 1495/3`

Vậy, `x = 1495/3`

`c,`

`390 \div (5x - 5) = 39`

`=> 5x - 5 = 390 \div 39`

`=> 5x - 5 = 10`

`=> 5x = 10 + 5`

`=> 5x = 15`

`=> x = 15 \div 5`

`=> x = 3`

Vậy, `x = 3`

`d,`

`34x - 14x = 200`

`=> (34 - 14)x = 200`

`=> 20x = 200`

`=> x = 200 \div 20`

`=> x = 10`

Vậy, `x = 10.`

`@` `\text {Kaizuu lv uuu}`

a) \(x-280:35=554\)

\(x-8=554\)

\(x=562\)

b) \(x+\left(5:3\right)=500\)

\(x+\dfrac{5}{3}=500\)

\(x=500-\dfrac{5}{3}=\dfrac{1500}{3}-\dfrac{5}{3}=\dfrac{1495}{3}\)

c) \(390:\left(5x-5\right)=39\)

\(5x-5=\dfrac{39}{390}=\dfrac{1}{10}\)

\(5x=5+\dfrac{1}{10}\)

\(5x=\dfrac{51}{10}\)

\(x=\dfrac{51}{10}.\dfrac{1}{5}=\dfrac{51}{50}\)

d) \(34x-14x=200\)

\(20x=200\)

\(x=200:20=10\)

hình như thứ tự có hơi..... Mình khó hiểu để ???

biết bài nào thì giúp mình bài đó nha, 0 phải làm hết đâu

b: =>(x+1)^2=36

=>x+1=6 hoặc x+1=-6

=>x=5 hoặc x=-5

e: 5x=4y

=>x/4=y/5

mà y-x=7

nên x/4=y/5=(y-x)/(5-4)=7

=>x=28; y=35

d: =>5^2x+7=25

=>2x+7=2

=>2x=-5

=>x=-5/2

g: =>2x^2=50

=>x^2=25

=>x=5 hoặc x=-5

`3 xx (x/4 + x/28 + x/70 + x/130) =60/13`

`x/4 + x/28 + x/70 + x/130 = 20/13`

`x xx (1/4+1/28+1/70+1/130)=20/13`

`x xx 4/13= 20/13`

`x=5`

`3 xx (x/4 + x/28 + +x/70 +x/130) =60/13`

`x xx (1/4 + 1/28 + +1/70 +1/130)=20/13

`x xx 4/13 = 20/13`

`x=5`

ta có : \(\dfrac{392-x}{32}+\dfrac{390-x}{34}+\dfrac{388-x}{36}+\dfrac{386-x}{38}\)+\(\dfrac{384-x}{40}=-5\)

\(\Leftrightarrow\)\(\dfrac{392-x}{32}+1+\dfrac{390-x}{34}+1+\dfrac{388-x}{36}+1\)+\(\dfrac{384-x}{40}+1=0\)

\(\Leftrightarrow\)\(\dfrac{424-x}{32}+\dfrac{424-x}{34}+\dfrac{424-x}{36}+\dfrac{424-x}{38}+\dfrac{424-x}{40}=0\)\(\Leftrightarrow\left(424-x\right)\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\right)=0\)

\(\Leftrightarrow x=424\)(vì \(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\ne0\))

Vậy tập nghiệm của phương trình là s=\(\left\{424\right\}\)

Giải:

a) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=30^3=2700\)

b) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=30^4=810000\)

c) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^2=8^2=64\)

Đáp số: a) 2700; b) 810000; c) 64.

Chúc bạn học tốt!!!

a, \(\left\{{}\begin{matrix}x+y=4\\\left(x^2+y^2\right)\left(x^3+y^3\right)=280\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(x^2+y^2\right)\left(x^2+y^2-xy\right)=70\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(16-2xy\right)\left(16-3xy\right)=70\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\3x^2y^2-40xy+93=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left[{}\begin{matrix}xy=\dfrac{31}{3}\\xy=3\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=4\\xy=\dfrac{31}{3}\end{matrix}\right.\)

Phương trình này vô nghiệm

Vậy hệ đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(1;3\right);\left(3;1\right)\right\}\)

b, ĐK: \(xy>0\)

\(\left\{{}\begin{matrix}\sqrt{\dfrac{2x}{y}}+\sqrt{\dfrac{2y}{x}}=3\\x-y+xy=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x}{y}+\dfrac{2y}{x}+4=9\\x-y+xy=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2+y^2\right)=5xy\\x-y+xy=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(x-2y\right)=0\\x-y+xy=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}2x=y\\x=2y\end{matrix}\right.\\x-y+xy=3\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}y=2x\\x-y+xy=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2x\\2x^2-x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2x\\\left(x+1\right)\left(2x-3\right)=0\end{matrix}\right.\)

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

TH2: \(\left\{{}\begin{matrix}x=2y\\x-y+xy=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2+y-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

Bài làm :

\(a,\left[\left(250-25\right):15\right]:x=\left(450-60\right):130\)

\(\Rightarrow\left(225:15\right):x=390:130\)

\(\Rightarrow15:x=3\)

\(\Rightarrow x=15:3\)

\(\Rightarrow x=5\)

\(b,390-\left(x-7\right)=169:13\)

\(\Rightarrow390-\left(x-7\right)=13\)

\(\Rightarrow x-7=390-13\)

\(\Rightarrow x-7=377\)

\(\Rightarrow x=377+7\)

\(\Rightarrow x=384\)

\(c,x-6:2-\left(48-24\right):2:6-3=0\)

\(\Rightarrow x-6:2-24:2:6=3\)

\(\Rightarrow x-3-2=3\)

\(\Rightarrow x=3+2+3\)

\(\Rightarrow x=8\)

Học tốt nhé

a) ((250-25):15):x=(450-60):130

(225:15):x=390:130

15:x=3

x=15:3=5

b) 390-(x-7)=169:13

390-x+7=13

x+7=390-13=377

x=377-7=370

c)x-6:2-(48-24):2:6-3=0

x-3-24:2:6-3=0

x-3-2-3=0

x-8=0

x=8

tk cho mk :>>

9) Theo bài, ta có : 5x = 4y

\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}\)

Mà y - x = 7

Theo tính chất dãy tỉ số bằng nhau, ta có :

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{y-x}{5-4}=\dfrac{7}{1}=7\)

Do đó : \(\dfrac{x}{4}=7\Rightarrow x=7.4=28\)

\(\dfrac{y}{5}=7\Rightarrow y=7.5=35\)

Vậy x = 28 ; y = 35

4) \(\left(\dfrac{1}{2}\right)^x\) = \(\dfrac{1}{32}\)

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

\(\Rightarrow x=5\)

Vậy x = 5

Nhớ tick cho mk nha !!!

3) \(2^x\) = 8

\(2^x\) = \(2^3\)

\(\Rightarrow x=3\)

Vậy x = 3