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15 tháng 12 2023

k: \(\left(4x-16\right)\left(-72+9x\right)=0\)

=>\(4\cdot\left(x-4\right)\cdot9\left(x-8\right)=0\)

=>\(36\left(x-4\right)\left(x-8\right)=0\)

=>\(\left(x-4\right)\left(x-8\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=8\end{matrix}\right.\)

m: \(\left(20+5x\right)\left(4x-8\right)=0\)

=>\(5\cdot\left(x+4\right)\cdot4\left(x-2\right)=0\)

=>\(\left(x+4\right)\left(x-2\right)=0\)

=>\(\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

n: \(\left(-4x+48\right)\left(2x-24\right)=0\)

=>\(-4\left(x-12\right)\cdot2\left(x-12\right)=0\)

=>\(\left(x-12\right)^2=0\)

=>x-12=0

=>x=12

o: \(\left(4x+16\right)\left(-2x+20\right)\left(-40+x\right)=0\)

=>\(4\cdot\left(x+4\right)\cdot\left(-2\right)\left(x-10\right)\left(x-40\right)=0\)

=>\(\left(x+4\right)\left(x-10\right)\left(x-40\right)=0\)

=>\(\left[{}\begin{matrix}x+4=0\\x-10=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=10\\x=40\end{matrix}\right.\)

p: \(\left(-5x+40\right)\left(-x+2023\right)\left(2x-2\right)=0\)

=>\(-5\left(x-8\right)\cdot\left(-1\right)\cdot\left(x-2023\right)\cdot2\left(x-1\right)=0\)

=>\(\left(x-8\right)\left(x-2023\right)\left(x-1\right)=0\)

=>\(\left[{}\begin{matrix}x-8=0\\x-1=0\\x-2023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\x=2023\end{matrix}\right.\)

q: \(2024x\left(4x-8\right)\left(5+5x\right)=0\)

=>\(x\cdot4\left(x-2\right)\cdot5\left(x+1\right)=0\)

=>\(x\left(x-2\right)\left(x+1\right)=0\)

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

r: \(-4x\left(3x+9\right)\left(2x-16\right)=0\)

=>\(-4x\cdot3\left(x+3\right)\cdot2\left(x-8\right)=0\)

=>\(x\left(x+3\right)\left(x-8\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x+3=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=8\end{matrix}\right.\)

s: \(\left(-100+5x\right)\left(2x-10\right)\left(6x+6\right)=0\)

=>\(5\cdot\left(x-20\right)\cdot2\left(x-5\right)\cdot6\left(x+1\right)=0\)

=>\(\left(x-20\right)\left(x-5\right)\left(x+1\right)=0\)

=>\(\left[{}\begin{matrix}x-20=0\\x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=5\\x=-1\end{matrix}\right.\)

t: \(\left(-2x+4\right)\left(2x+16\right)\cdot\left(7-x\right)=0\)

=>\(-2\left(x-2\right)\cdot2\left(x+8\right)\cdot\left(-1\right)\cdot\left(x-7\right)=0\)

=>\(\left(x-2\right)\left(x+8\right)\left(x-7\right)=0\)

=>\(\left[{}\begin{matrix}x-2=0\\x-7=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\\x=7\end{matrix}\right.\)

15 tháng 12 2023

a: \(\left(x+10\right)\left(x-5\right)=0\)

=>\(\left[{}\begin{matrix}x+10=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=5\end{matrix}\right.\)

b: \(\left(2x+10\right)\left(4+x\right)=0\)

=>\(\left[{}\begin{matrix}2x+10=0\\4+x=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-4\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)

c: \(\left(4x+20\right)\left(12x-24\right)=0\)

=>\(\left[{}\begin{matrix}4x+20=0\\12x-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-20\\12x=24\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

d: \(\left(x-2024\right)\left(4x+4\right)=0\)

=>\(\left[{}\begin{matrix}x-2024=0\\4x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2024\\4x=-4\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-1\\x=2024\end{matrix}\right.\)

e: \(\left(2x-6\right)\left(7+x\right)=0\)

=>\(\left[{}\begin{matrix}2x-6=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\x=-7\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)

g: (4x+8)(6-x)=0

=>\(\left[{}\begin{matrix}4x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x=6\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\)

h: (2x+2)(4x-8)=0

=>2(x+1)*4*(x-2)=0

=>(x+1)(x-2)=0

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

i: (2x-2024)(8x-16)=0

=>\(2\left(x-1012\right)\cdot8\cdot\left(x-2\right)=0\)

=>\(\left(x-1012\right)\left(x-2\right)=0\)

=>\(\left[{}\begin{matrix}x-1012=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1012\\x=2\end{matrix}\right.\)

20 tháng 12 2023

e: \(\left(-4156+2021\right)-\left(119+2021-4156\right)\)

\(=-4156+2021-119-2021+4156\)

\(=\left(-4156+4156\right)+\left(2021-2021\right)-119\)

=0+0-119

=-119

g: \(315\cdot75-\left(15\cdot100-315\cdot25\right)\)

\(=315\cdot75-15\cdot100+315\cdot25\)

\(=315\left(75+25\right)-15\cdot100\)

\(=315\cdot100-15\cdot100=300\cdot100=30000\)

h: \(\left(-489\right)\cdot125-\left(125\cdot11-500\cdot25\right)\)

\(=-489\cdot125-125\cdot11+500\cdot25\)

\(=125\left(-489-11\right)+500\cdot25\)

\(=125\cdot\left(-500\right)+500\cdot25\)

\(=500\left(-125+25\right)\)

\(=500\cdot\left(-100\right)=-50000\)

Bài 2:

a: \(-415-3\left(2x-1\right)^2=-490\)

=>\(3\left(2x-1\right)^2+415=490\)

=>\(3\left(2x-1\right)^2=75\)

=>\(\left(2x-1\right)^2=25\)

=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

29 tháng 7 2018

Ta có :  \(x-y=4\Rightarrow x=4+y\)

Vì  7x5y1 chia hết cho 3 

\(\Rightarrow7+4+y+5+y+1⋮3\)

\(\Leftrightarrow17+2y⋮3\)

\(\Rightarrow2y=\left\{1;4;7;10;13;16;19\right\}\)

\(\Rightarrow y=\left\{\frac{1}{2};2;\frac{7}{2};5;\frac{13}{2};8;\frac{19}{2}\right\}\)  Mà \(y\in N\)\(\Rightarrow y=\left\{2;5;8\right\}\)

\(\Rightarrow x=\hept{\begin{cases}2+4=6\\5+4=9\\8+4=12\end{cases}\Rightarrow x=\left\{6;9\right\}}\) 

\(y=\left\{2;5\right\}\)

( Vì x;y có 1 chữ số <=> là số tự nhiên nhỏ hơn 10 )

Vậy ta có các cặp x ; y  :   ( 2 ; 5 )  ( 5 ; 8 )

Câu 2: 

1: \(\Leftrightarrow x\cdot\dfrac{7}{2}=\dfrac{9}{2}+3=\dfrac{15}{2}\)

hay x=15/7

2: \(\Leftrightarrow x=\dfrac{5}{2}\cdot\dfrac{8}{5}=4\)

3: \(\Leftrightarrow x=\dfrac{-11\cdot10}{5}=-11\cdot2=-22\)

4: =>2x=90

hay x=45

10 tháng 12 2021

Answer:

Bài 3:

\(334-\left(-75\right)+[\left(-175\right)-34]\)

\(=334+75-175-34\)

\(=\left(334-34\right)+\left(75-175\right)\)

\(=300+\left(-100\right)\)

\(=200\)

\(68-\left(-45\right)-[368-\left(-145\right)]\)

\(=68+45-[368+145]\)

\(=\left(68-368\right)+\left(45-145\right)\)

\(=\left(-300\right)+\left(-100\right)\)

\(=-400\)

Bài 4:

\(-17-\left(x-5\right)=-52\)

\(\Rightarrow x-5=-17-\left(-52\right)\)

\(\Rightarrow x-5=35\)

\(\Rightarrow x=40\)

\(3\left(7-x\right)+\left(-12\right)=-75\)

\(\Rightarrow3\left(7-x\right)=-75-\left(-12\right)\)

\(\Rightarrow3\left(7-x\right)=-63\)

\(\Rightarrow7-x=-21\)

\(\Rightarrow x=28\)

Bài 8:

a: \(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)

b: \(B=2\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)

\(=2\left(\dfrac{1}{15}-\dfrac{1}{90}\right)\)

\(=2\cdot\dfrac{5}{90}=\dfrac{10}{90}=\dfrac{1}{9}\)

c: \(C=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

\(=3\cdot\dfrac{24}{200}=\dfrac{72}{200}=\dfrac{9}{25}\)