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a: \(=\dfrac{17}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{7}{2}\)
\(=\dfrac{35}{8}+\dfrac{8}{8}-\dfrac{37}{100}-\dfrac{128}{100}\)
\(=\dfrac{43}{8}-\dfrac{165}{100}=\dfrac{149}{40}\)
b: \(=\left(\dfrac{22\cdot26+3\cdot10-65}{130}\right):\left(\dfrac{4\cdot22-2\cdot26+3\cdot143}{286}\right)\)
\(=\dfrac{537}{130}\cdot\dfrac{286}{465}=\dfrac{1969}{775}\)
| x | 7 | 9 | |||
| x2 | 49 | 81 | |||
| x2-49 | - | 0 | + | + | + |
| x2-81 | - | - | - | 0 | + |
| A | + | 0 | - | 0 | + |
dựa vào bảng ta có khi 7<x<9 thì A<0 vậy 7<x<9
b, ta có : \(\frac{2015}{1}\)+\(\frac{2014}{2}\)+\(\frac{2013}{3}\)+......+\(\frac{1}{2015}\)
=1+1+1+1......+1+\(\frac{2014}{2}\)+\(\frac{2013}{3}\)+.......+\(\frac{1}{2015}\)
(2015 số 1)
=1+(1+\(\frac{2014}{2}\))+(1+\(\frac{2013}{3}\))+........+(1+\(\frac{1}{2015}\))
=\(\frac{2016}{2016}\)+\(\frac{2016}{2}\)+\(\frac{2016}{3}\)+.........+\(\frac{2016}{2015}\)
=2016(\(\frac{1}{2016}\)+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.........+\(\frac{1}{2015}\))
\(\)\(\left|x-1,5\right|+\left|2,5-x\right|=0\)
Với mọi \(x\in R\) thì:
\(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix} \left|x-1,5\right|=0\\ \left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)
Khi đó không tồn tại giá trị x
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\sqrt{\dfrac{1}{6}}\\x+\dfrac{1}{2}=-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\sqrt{\dfrac{1}{6}}\\x=\dfrac{1}{2}-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\)
\(\sqrt{\dfrac{1}{6}=?}\)
mk ko hiểu Linh Nguyễn
mk chưa hk đến căn
\(1\)) \(70:\frac{4x+720}{x}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4x+720}{x}=70:\frac{1}{2}\)
\(\Leftrightarrow\frac{4x+720}{x}=140\)
\(\Leftrightarrow\left(4x+720\right):x=140\)
\(\Leftrightarrow4x+720=140.x\)
\(\Leftrightarrow4x-140x=-720\)
\(\Leftrightarrow x.\left(-136\right)=-720\)
\(\Leftrightarrow x=-720:\left(-136\right)\)
\(\Leftrightarrow x=\frac{90}{17}\)
\(2\)) Mình đang nghĩ
a) \(2^x+5=21\)
\(2^x=21-5\)
\(2^x=16\)
\(2^x=2^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
a ) \(2^x+5=21\)
\(2^x=21-5\)
\(2^x=16\)
\(2^x=2^4\)
\(\Rightarrow x=4\)
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
b)\(\left(2x-3\right)^3=343\)
\(\left(2x-3\right)^3=7^3\)
\(2x-3=7\)
\(2x=7+3\)
\(2x=10\)
\(x=10:2\)
\(x=5\)
a) Ta có: \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
<=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
<=> \(x=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
Vậy x=5/6
b)\(\left(2x-3\right)^3=343\)
<=>\(2x-3=\sqrt[3]{343}=7\)
<=> 2x=10 <=> x=5
c) \(\left(\frac{1}{3}\right)^{2x}+1=\frac{1}{7}\)
<=>\(\left(\frac{1}{3}\right)^{2x}=\frac{-6}{7}\)
<=> \(\left(\frac{1}{3^x}\right)^2=-\frac{6}{7}\)(vô lí vì \(\left(\frac{1}{3^x}\right)^2\ge0\))
Vậy ko tìm được x thỏa mãn.
d)\(\left(2x-3\right)^2=9\)
=>\(\left[\begin{array}{nghiempt}2x-3=3\\2x-3=-3\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=3\\x=0\end{array}\right.\)
Vậy x=3 hoặc x=0.
e) \(\left(x-3\right)^6=\left(x-3\right)^7\)
<=> \(\left(x-3\right)^7-\left(x-3\right)^6=0\)
<=> \(\left(x-3\right)^6\left(x-3-1\right)=0\)
<=>\(\left(x-3\right)^6\left(x-4\right)=0\)
<=> \(\left[\begin{array}{nghiempt}x-3=0\\x-4=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=3\\x=4\end{array}\right.\)
Vậy x \(\in\left\{3;4\right\}\)
cảm ơn bn