Bài 1:Tìm x
a) 71 - (33 + x) = 26
b) 97 - (64 - x) = 44
c) x - 36 : 18 = 12
d) 3636 : (12 . x- 91) = 36
e) ( x : 23 + 45) . 67 = 8911
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e)
\(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ 1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ 1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 2-\dfrac{1}{100}< 2\)
\(x^2\) + 6\(x\) + 9 = 25
\(x^2\) + 6\(x\) + 9 - 25 = 0
\(x^2\) + 6\(x\) + (9 - 25) = 0
\(x^2\) + 6\(x\) - 16 = 0
\(x^2\) - 2\(x\) + 8\(x\) - 16 = 0
(\(x^2\) - 2\(x\)) + (8\(x\) - 16) = 0
\(x\)(\(x\) - 2) + 8(\(x-2\)) = 0
(\(x\) - 2)(\(x\) + 8) = 0
\(\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
Vậy \(x\) \(\in\) {2; - 8}
a)
\(3^x=81\\ 3^x=3^4\\ x=4\)
b)
\(\left(3x-5\right)^2=49\\ \left(3x-5\right)^2=7^2\)
TH1: 3x - 5 = 7
3x = 7 + 5
3x = 12
x = 12 : 3
x = 4
TH2: 3x - 5 = -7
3x = -7 + 5
3x = -2
x = -2/3
c) 68 - ? = 36
d)
\(\left(7-2x\right)^3=27\\ \left(7-2x\right)^3=3^3\\ 7-2x=3\\ 2x=7-3\\ 2x=4\\ x=\dfrac{4}{2}\\ x=2\)
Đặt:
\(A=\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{25^{10}}\\ A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{\left(5^2\right)^{10}}\\ A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{20}}\\ 5A=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{19}}\\ 5A-A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{19}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{20}}\right)\\4A =1-\dfrac{1}{5^{20}}\\ 4A=\dfrac{5^{20}-1}{5^{20}}\\ A=\dfrac{5^{20}-1}{4\cdot5^{20}}\)
$(x+1)+(x+2)+...+(x+50)=1175$
$\Rightarrow (x+x+...+x)+(1+2+...+50)=1175$ (1)
Số các số hạng x trong tổng đó là: $(50-1):1+1=50$ (số)
Tổng các số từ 1 đến 50 là:
$(50+1).50:2=1275
Khi đó, (1) trở thành:
$50x+1275=1175$
$\Rightarrow 50x=1175-1275$
$\Rightarrow 50x=-100$
$Rightarrow x=-100:50=-2$
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+50\right)=1175\)
\(50x+\left(1+2+...+50\right)=1175\)
\(50x+\dfrac{50\cdot\left(50+1\right)}{2}=1175\)
\(50x+1275=1175\)
\(50x=1175-1275\)
\(50x=-100\)
\(x=-100:50\)
\(x=-2\)
Vậy...
sửa \(\left(x+1\right)+\left(x+2\right)+...+\left(x+50\right)=1175\)
\(\Leftrightarrow50x+1275=1175\Leftrightarrow x=-2\)
\(a)71-\left(33+x\right)=26\\ 33+x=71-26=45\\ x=45-33\\ x=12\\ b)97-\left(64-x\right)=44\\ 64-x=97-44\\ 64-x=53\\ x=64-53\\ x=11\\ c)x-36:18=12\\ x-2=12\\ x=2+12\\ x=14\\ d)3636:\left(12\cdot x-91\right)=36\\ 12\cdot x-91=3636:36\\ 12\cdot x-91=101\\ 12\cdot x=101+91\\ 12\cdot x=192\\ x=\dfrac{192}{12}\\ x=16\\ e)\left(x:23+45\right)\cdot67=8911\\ x:23+45=8911:67\\ x:23+45=133\\ x:23=133-45=88\\ x=88\cdot23\\ x=2024\)
Cứu voiii