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\(a+a.2+a.3+...+a.49+a.50=12750\)
\(\Leftrightarrow a\left(1+2+3+...+49+50\right)\)
\(\Leftrightarrow1275a=12750\)\
\(\Leftrightarrow a=\frac{12750}{1275}=10\)
Vậy \(a=10\)
\(\left(2x-3\right)^2=7^2\)
\(2x-3=7\)
\(2x=10\)
\(x=5\)
Vậy x=5
a: \(\left(2x-3\right)^2-49=0\)
\(\Leftrightarrow\left(2x+4\right)\left(2x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
1.Tính nhanh nếu có thể:
a) 22 + 23 + 89 + 77
= ( 77 + 23 ) + 22 + 89
= 100 + 22 + 89
= 122 + 89
= 211
b) 35 . 15 + 15 . 65
= 15 . ( 35 + 65 )
= 15 . 100
= 1500
c) 7^2 - 36 : 3^2
= 7^2 - 36 : 9
= 7^2 - 4
= 49 - 4
= 45
d) 476 - {5 . [409 - (8 . 3 - 21)2] - 1724}
= 476 - {5 . [409 - (24 - 21)^2] - 1724}
= 476 - {5 . [409 - (3^2)] - 1724}
= 476 - {5 . [409 - 9 ] - 1724}
= 476 - {5. 400 - 1724}
= 476 - {2000 - 1724}
= 476 - 276
= 200
a) x ( 3 y + 7 ) - 5 ( 3 y + 7 ) + 35 = 104
=> ( x - 5 ) ( 3 y + 7 ) = 104 - 35 = 69
ta có bảng
..............
a)\(3xy+7x-15y=104\)
\(=>3xy+7x-15y-104=0\)
\(=>3xy-15y+7x-35-69=0\)
=>\(3y\left(x-5\right)+\left(7x-35\right)=69\)
\(=>3y\left(x-5\right)+7\left(x-5\right)=69\)
=>\(\left(3y+7\right)\left(x-5\right)=69\)
Ta có\(69=\pm1.\pm69=\pm3.\pm23=\pm69.\pm1=\pm23.\pm3\)
tự phân TH ra làm nốt nha
a: \(2\left(x-51\right)=2\cdot2^3+20\)
=>\(2\left(x-51\right)=2^4+20=36\)
=>x-51=36/2=18
=>x=18+51=69
b: \(2x-49=5\cdot3^2\)
=>\(2x-49=5\cdot9=45\)
=>2x=45+49=94
=>x=94/2=47
c: \(\left[\left(8x-12\right):4\right]\cdot3^3=3^6\)
=>\(\left[4\cdot\dfrac{\left(2x-3\right)}{4}\right]=3^3\)
=>\(2x-3=3^3=27\)
=>2x=3+27=30
=>x=30/2=15
d: \(2^{x+1}-2^2=32\)
=>\(2^{x+1}=32+2^2=32+4=36\)
=>\(x+1=log_236\)
=>\(x=log_236-1\)
e: \(\left(x^3-77\right):4=5\)
=>\(x^3-77=20\)
=>\(x^3=77+20=97\)
=>\(x=\sqrt[3]{97}\)
Bài làm
b) Ta có: b + c = 49 => b = 49 - c
c + a = 10 => a = 10 - c
Thay a và b vừa tìm được vào a + b = -21 ta được:
49 - c + 10 - c = -21
59 - 2c = -21
=> 2c = 80
=> c = 40
Thay c = 40 và b + c = 49 ta được:
b + 40 = 49 => b = 9
Thay c = 40 và c + a = 10 ta được:
40 + a = 10 => a = -30
Vậy a = -30; b = 9; c = 40
a)
x − 5 12 − 4 9 = − 13 18 x = − 13 18 + 5 12 + 4 9 x = − 26 36 + 15 36 + 16 36 x = 5 36
b)
2 − x + 3 7 = 9 − 21 2 − x − 3 7 = − 3 7 − x = − 3 7 + 3 7 − 2 − x = − 2 ⇒ x = 2
a) \(\Rightarrow\left(2x-3\right)^2=49\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c) \(\Rightarrow x\left(x-5\right)+2\left(x-5\right)=0\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a, ⇒ (2x - 3)2 = 49
⇒ (2x - 3)2 = \(\left(\pm7\right)^2\)
⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0
⇒ (x - 5).(2x + 7) = 0
⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c, ⇒ x2 - 5x + 2x - 10 = 0
⇒ (x2 - 5x) + (2x - 10) = 0
⇒ x.(x - 5) +2.(x - 5) = 0
⇒ (x - 5).(x + 2)=0
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
A x 2 = 37240 + 49346
A x 2 = 86586
A = 86 586 : 2
A = 43293
a x 2 = 37 240 + 49 346
a x 2 = 86586
a = 86586 : 2
a = 43293