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\(445-52+155+452\\ =\left(445+155\right)+\left(452-52\right)\\ =600+4000\\ =1000\\ 102\cdot55-102\cdot73+51\cdot36\\ =102\cdot\left(55-73\right)+51\cdot36\\ =102\cdot\left(-18\right)+102\cdot18\\ =0\\ 4\cdot5\cdot2\cdot25\cdot5-1000\\ =5000-1000\\ =4000\\ 8782-291\cdot13\\ =4999\\ 254+\left\{38\cdot\left[\left(42-16\right):2\right]\right\}\\ =254+\left[38\cdot\left(26:2\right)\right]\\ =254+\left(38\cdot13\right)\\ =748\\ \left(2791\cdot34+7882:14\right)\cdot0+1510-510:2\\ =1510-255\\ =1255\)
a. \(445-52+155+452=\left(445+155\right)+\left(452-52\right)=600+400=1000\)
b. \(102\cdot55-102\cdot73+51\cdot36=102\left(55-73+18\right)=102\cdot0=0\)
c.\(4\cdot5\cdot2\cdot25\cdot5-1000=4\cdot25\cdot5\cdot5\cdot2-1000=100\cdot50-1000=5000-1000=4000\)
d. \(8782-291\cdot13=8782-3783=4999\)
e. \(254+\left\{38\left[\left(42-16\right)\div2\right]\right\}=254+\left\{30\cdot13\right\}=254+494=748\)
f. \(\left(2791\cdot34+7882\div14\right)\cdot0+1510-510\div2=0+1510-255=1255\)\
a,(72000+18000)-(3x+3000)=12000
90000-(3x+3000)=12000
3x+3000=90000-12000
3x+3000=78000
3x=78000-3000
3x=75000
x=75000:3
x=25000
b,[3.(x+2):7]:4=120
3.(x+2):7=120:4
3.(x+2):7=30
3.(x+2)=30.7
3.(x+2)=210
x+2=210:3
x+2=70
x=70-2
x=68
( x + 70 ) : 14 - 40 = (32 . 15 ) : 2
( x + 70 ) : 14 - 40 = 240
( x + 70 ) : 14 = 240 + 40 = 280
x + 70 = 280 x 14 = 3920
x = 3920 - 70
x = 3850
Vậy x = 3850.
(x+70):14-40=(32.15):2
(x+70):14-40=480;2
(x+70):14-40=240
(x+70);14=240+40
(x+70):14=280
(x+70)=280.14
(x+70)=3360
x=3360-70
x=3290
a)\(5^{2x-3}-2\cdot5^2=5^2\cdot3\)
\(5^{2x-3}-2\cdot25=75\)
\(5^{2x-3}-50=75\)
\(5^{2x-3}=125\)
\(125=5^3\)
\(5^3=5^{\left(3+3\right):2}=5^3\Rightarrow x=3\)
Vậy \(x=3\)
b)\(\frac{2}{9}\cdot\left(5x+1\right):2-\frac{1}{18}=\frac{5}{36}\)
\(=\frac{2}{9}\cdot\left(5x+1\right):2=\frac{5}{36}+\frac{1}{18}\)
\(=\frac{2}{9}\cdot\left(5x+1\right)=\frac{7}{36}\cdot2\)
\(5x+1=\frac{7}{18}:\frac{2}{9}\)
\(x=\left(\frac{7}{4}-1\right):5=\frac{3}{20}\)
\(\Rightarrow x=\frac{3}{20}\)
a, \(5^{2x-3}-2.5^2=5^2.3\)
\(\Leftrightarrow5^{2x-3}-2.25=25.3\)
\(\Leftrightarrow5^{2x-3}-50=75\)
\(\Leftrightarrow5^{2x-3}=125\)
\(\Leftrightarrow5^{2x-3}=5^3\)
\(\Leftrightarrow2x-3=3\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b, \(\frac{2}{9}.\left(5x+1\right):2-\frac{1}{18}=\frac{5}{36}\)
\(\Leftrightarrow\frac{2}{9}.\left(5x+1\right):2=\frac{7}{36}\)
\(\Leftrightarrow\frac{2}{9}.\left(5x+1\right)=\frac{7}{36}.2\)
\(\Leftrightarrow\frac{2}{9}.\left(5x+1\right)=\frac{7}{18}\)
\(\Leftrightarrow5x+1=\frac{7}{18}:\frac{2}{9}\)
\(\Leftrightarrow5x+1=\frac{7}{18}.\frac{9}{2}\)
\(\Leftrightarrow5x+1=\frac{7}{4}\)
\(\Leftrightarrow5x=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}:5=\frac{3}{4}.\frac{1}{5}=\frac{3}{20}\)
Theo đề ta có a=5k+2
b=5q+3
13a+11b=13(5k+2)+11(5q+3)=65k+26+55q+33=(65k+55q)+59
Ta có 65k+55q chia hết cho 5 vì mỗi số hạng đều chia hết cho 5
59 chia 5 dư 4
Vậy 13a+11b chia 5 dư 4
\(5\cdot x+10\cdot9=990\)
\(\Rightarrow5\cdot x+90=990\)
\(\Rightarrow5\cdot x=900\)
\(\Rightarrow x=\dfrac{900}{5}\)
\(\Rightarrow x=180\)
_____________
\(1045:\left[215-\left(3\cdot x-24\right)\right]=5\)
\(\Rightarrow215-\left(3\cdot x-24\right)=1045:5\)
\(\Rightarrow215-\left(3\cdot x-24\right)=209\)
\(\Rightarrow3\cdot x-24=215-209\)
\(\Rightarrow3\cdot x-24=6\)
\(\Rightarrow3\cdot x=30\)
\(\Rightarrow x=10\)
`@` `\text {Ans}`
`\downarrow`
`5.x + 10.9 = 990`
`\Rightarrow 5x + 90 = 990`
`\Rightarrow 5x = 990 - 90`
`\Rightarrow 5x = 900`
`\Rightarrow x = 900 \div 5`
`\Rightarrow x = 180`
Vậy, `x = 180`
\(1045 \div [ 215 - (3 . x - 24 ) ] = 5\)
`\Rightarrow 1045 \div (215 - 3x + 24) = 5`
`\Rightarrow 191 + 3x = 1045 \div 5`
`\Rightarrow 191 + 3x = 209`
`\Rightarrow 3x = 209 - 191`
`\Rightarrow 3x =18`
`\Rightarrow x = 18 \div 3`
`\Rightarrow x = 6`
Vậy, `x = 6.`
\(\left(3x-2\right)^2=14-2\cdot5^2\)
=>\(\left(3x-2\right)^2=14-2\cdot25=14-50=-36\)
mà \(\left(3x-2\right)^2>=0\forall x\)
nên \(x\in\varnothing\)
\(\left(3x-2\right)^2=14-2.5^2\)
\(\Rightarrow\left(3x-2\right)^2=14-2.25\)
\(\Rightarrow\left(3x-2\right)^2=14-50\)
\(\Rightarrow\left(3x-2\right)^2=-36\)
Vì \(\left(3x-2\right)^2\ge0\) với mọi \(x\)
\(\Rightarrow x\in\left\{\varnothing\right\}\)