12 + 3.{90 : [39 - (23 - 5)2]}
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b: \(=9\left(148-48\right)=9\cdot100=900\)
c: \(=307-\left[\left(180-160\right):4+9\right]:2\)
\(=307-\left(5+9\right):2=307-7=300\)
d: \(=12+3\cdot\left\{90:\left[39-3^2\right]\right\}=12+3\cdot\left(90:30\right)=12+3\cdot3=21\)
a) 45.37 + 45.63 - 100
\( 45.(37 + 63) – 100 \)
\(45.100 – 100\)
\(100.(45 – 1) \)
\( 100 . 44 = 4400\)
Ta có A = \(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
= \(\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)
= \(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
= \(\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\)
B = \(\dfrac{\dfrac{2}{29}-\dfrac{2}{39}+\dfrac{2}{49}}{\dfrac{23}{29}-\dfrac{23}{39}+\dfrac{23}{49}}=\dfrac{2\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}{23\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}=\dfrac{2}{23}\)
Lại có \(\dfrac{2}{23}>\dfrac{2}{24}=\dfrac{1}{12}\) hay A < B
Vậy A < B
a) \(25-\left(25-x\right)=12+\left(42-65\right)\)
\(25-\left(25-x\right)=12+\left(-23\right)\)
\(25-\left(25-x\right)=-11\)
\(25-x=25-\left(-11\right)\)
\(25-x=36\)
\(x=25-36\)
\(x=-11\)
Làm 2 câu dài cho bạn nhá! Mấy câu ngắn tương tự!
a, \(25-\left(25-x\right)=12+\left(42-65\right)\)
\(\Rightarrow25-25+x=12+42-65\)
\(\Rightarrow x=-11\)
b, \(215+\left(-38\right)-54+90-\left(-48+x\right)=0\)
\(\Rightarrow215-38-54+90+48-x=0\)
\(\Rightarrow x=261\)
m, \(\left|x\right|-5=12\)
\(\Rightarrow\left|x\right|=17\Rightarrow\left\{{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)
Chúc bạn học tốt!!!
\(=12+3\cdot\left\{90:\left[39-18\cdot2\right]\right\}\)
\(=12+3\cdot90:3=12+90=102\)
12+3.{90:[39-(2³-5)²]}
=12+3.{90:[39-9]}
=12+3.{90:30}
=12+3.3
=12+9
=21