Bài 3:

a: Ta có: \(23\left(42-x\right)=23\)

\(\Leftrightarrow42-x=1\)

hay x=41

b: Ta có: 15(x-3)=30

nên x-3=2

hay x=5

Bài 1: 

a: 32+89+68=100+89=189

b: 64+112+236=300+112=412

c: \(1350+360+650+40=2000+400=2400\)

98775 - 32 x 85

=98775 -2720

=96055

 67500 - 24 x 236

= 67500 -5664

=61836

 568 + 101598 : 287

= 568 +354

=922

6875 + 980 -180  

=7855 -180 

=7675

\(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\)

\(=\dfrac{7}{10}-\dfrac{1}{2}\)

= \(\dfrac{1}{5}\)

\(\dfrac{8}{11}+\dfrac{8}{33}x\dfrac{3}{4}\)

\(=\dfrac{8}{11}+\dfrac{2}{11}\)

\(=\dfrac{10}{11}\)

\(\dfrac{7}{9}x\dfrac{3}{14}:\dfrac{5}{8}\)

\(=\dfrac{1}{6}:\dfrac{5}{8}\)

\(=\dfrac{1}{6}x\dfrac{8}{5}\)

\(=\dfrac{8}{30}\)

\(=\dfrac{4}{15}\)

\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)

\(=\dfrac{5}{12}-\dfrac{7}{32}x\dfrac{16}{21}\)

\(=\dfrac{5}{12}-\dfrac{1}{6}\)

\(=\dfrac{5}{12}-\dfrac{2}{12}\)

\(=\dfrac{3}{12}=\dfrac{1}{4}\)

_{Tín hịu cầu kíu :>>}

132 - (236 - 432 - 576) = 132 - (-196 - 576) = 132 - (-772) = 904

55 - [132 - 2 x (45 - 42)²] + 7 = 55 - (132 - 2 x 3²) + 7 = 55 - (132 - 2 x 9) + 7 = 55 - (132 - 18) + 7 = 55 - 114 + 7 = -52

Hoc tot!!!

132-(236-432-576)=132-236+432+576=-104+432+576=328+576=904

55-[132-2(45-42)²]+7

= 55-[132-2*3²]+7

= 55-[132-2*9]+7

= 55-[132-18]+7

= 55-114+7=-52

(Chúc học tốt nha :) :) )

\(2,\)

\(a,20-\left[4^2+\left(x-6\right)\right]=90\)

\(\Rightarrow20-16-x+6=90\)

\(\Rightarrow10-x=90\)

\(\Rightarrow x=-80\)

Vậy: \(x=-80\)

\(b,\left(x+3\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)

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

Vậy: \(x\in\left\{-3;2\right\}\)

\(c,1000:\left[30+\left(2^x-6\right)\right]=3^2+4^2\left(x\in N\right)\)

\(\Rightarrow1000:\left(30+2^x-6\right)=25\)

\(\Rightarrow24+2^x=40\)

\(\Rightarrow2^x=16\)

\(\Rightarrow x=4\)

Vậy: \(x=4\)

:))))

\(2,\)

\(a,20-\left[42+\left(x-6\right)\right]=90\)

\(\Rightarrow20-42-x+6-90=0\)

\(\Rightarrow x=-106\)

Vậy: \(x=-106\)

\(b,\left(x+3\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)

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

Vậy: \(x\in\left\{-3;2\right\}\)

\(c,1000:\left[30+\left(2x-6\right)\right]=32+42\left(x\in N\right)\)

\(\Rightarrow1000:\left(30+2x-6\right)=74\)

\(\Rightarrow1000:\left(24+2x\right)=74\)

\(\Rightarrow24+2x=\dfrac{500}{37}\)

\(\Rightarrow2x=-\dfrac{388}{37}\)

\(\Rightarrow x=-\dfrac{194}{37}\)

Mà \(x\in N\)

\(\Rightarrow x\in\varnothing\)

Vậy: \(x\in\varnothing\)

a) => (x+32) - 17 = 42:2 = 21

=> x+32 = 21+17 = 38

=> x=38-32=6

b) => 61+(53-x) = 1785:17=105

=> 53-x = 105-61=44

=> x = 53-44 =9

c) => x^2 +54 -32 = 244:2 = 122

=> x^2 +22 = 122

=> x^2 = 122-2=100

=> x= 10 hoặc -10

giải oy pn **** giùm mk nka

a. X = 6

b. X = 9

c. X = 10

\(32-2\left(x+1\right)=42\)

\(\Rightarrow32-2x-1=42\)

\(\Rightarrow31-2x=42\)

\(\Rightarrow2x=31-42\)

\(\Rightarrow2x=-11\)

\(\Rightarrow x=\frac{-11}{2}\)

\(\Rightarrow x=-5,5\)

32-2.(x+1)=42

=>2.(x+1)=32-42

=>2.(x+1)=-10

=>x+1=-10:2

=>x+1=-5

=>x=-5-1

=>x=-6

\(32-2\left(x+1\right)=42\)

\(\Leftrightarrow2\left(x+1\right)=32-42\)

\(\Leftrightarrow2\left(x+1\right)=-10\)

\(\Leftrightarrow x+1=-10\div2\)

\(\Leftrightarrow x+1=-5\)

\(\Leftrightarrow x=-5-1\)

\(\Leftrightarrow x=-6\)

Vậy x=-6.

32-2.(x+1)=42

2.(x+1)=32-42

2(x+1)=-10

x+1=(-10):2

x+1=-5

x=-5-1

x=-6

32-2.(x+1)=42

2.(x+1)=42+32

2.(x+1)=74

x+1=74:2

x+1=37

x=37-1

x=36

chúc bạn hok tốt

\(32-2.\left(x+1\right)=42\)

\(2.\left(x+1\right)=32-42=-10\)

\(x+1=-10:2=-5\)

\(x=-5-1=-6\)

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

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

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)