Tính giá trị biểu thức:
a. 1485 x 58 + 24336 : 6
b. 99760 : 8 – 74088 : 7
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\(\text{a. 78,9 : 2,5 : 4. }\)
\(=31,56:4=7,89.\)
\(\text{ b. 56,75 + 8,4 : 4}.\)
\(=56,75+2,1=58,85.\)
\(\text{c. 23,6 x 100 + 2,9 : 0,1. }\)
\(=2360+29=2389.\)
\(\text{ d. 58 - 4,4 : 0,1}.\)
\(=58-44=14.\)
a) Với a = 3,05 thì ta có:
\(a\times2,46\) \(=3,05\times2,46=7,503\)
b) Với \(a=\dfrac{15}{8}\) thì ta có:
\(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):a\) \(=\left(\dfrac{5}{6}+\dfrac{7}{12}\right):\dfrac{15}{8}\) \(=\left(\dfrac{10}{12}+\dfrac{7}{12}\right)\times\dfrac{8}{15}\) \(=\dfrac{17}{12}\times\dfrac{8}{15}\) \(=\dfrac{34}{45}\)
a) Thay a=3,05 vào 2,46a, ta được:
\(2.46\cdot3.05=7.503\)
b) Thay \(a=\dfrac{15}{8}\) vào biểu thức \(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):a\), ta được:
\(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):\dfrac{15}{8}\)
\(=\dfrac{17}{12}\cdot\dfrac{8}{15}=\dfrac{34}{45}\)
|(5/8-5/14)+(3/8-9/14)|:4/7
=|(5/8+3/8)+(-5/14-9/14)|:4/7
=|1+(-1)|:4/7
=0
a, Thay x = 3 và y = -6 vào bt ta đc
\(5.3-4.\left(-6\right)=15-\left(-24\right)=39\\ b,\\ 2.\left(-2\right)^2-5.4=8-20=\left(-12\right)\\ c,\\ 5.\left(-1\right)^2+3.\left(-1\right)-1=5+\left(-3\right)-1=1\)
a) Thay x=3; y=-6
\(5x-4y=5.3-4.\left(-6\right)=15+24=39\)
b) Thay x=-2; y=4
\(2x^4-5y=2.\left(-2\right)^4-5.4=32-20=12\)
c, Thay x=0
\(5x^2+3x-1=5.0+3.0-1=-1\)
+) x=-1
\(5x^2+3x-1=5.\left(-1\right)^2+3.\left(-1\right)-1=5-3-1=1\)
+) \(x=\dfrac{1}{3}\)
\(5x^2+3x-1=5.\left(\dfrac{1}{3}\right)^2+3.\dfrac{1}{3}-1\)
\(=\dfrac{5}{9}+1-1=\dfrac{5}{9}\)
\(a^2-2a+6b+b^2=-10\\ \Leftrightarrow a^2-2a+1+b^2+6b+9=0\\ \Leftrightarrow\left(a-1\right)^2+\left(b+3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-3\end{matrix}\right.\)
Vậy \(\left(a;b\right)=\left(1;-3\right)\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ \Leftrightarrow xy+yz+zx=0\\ \Rightarrow\left\{{}\begin{matrix}xy+yz=-zx\\xy+zx=-yz\\yz+zx=-xy\end{matrix}\right.\)
Ta có:
\(A=\dfrac{xz+yz}{z^2}+\dfrac{xy+yz}{y^2}+\dfrac{xy+xz}{x^2}\\ =\dfrac{-xy}{z^2}+\dfrac{-xz}{y^2}+\dfrac{-yz}{x^2}\\ =-xyz\cdot\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)\\ =-xyz\cdot\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{xz}\right)\\ =0\)
\(a,256+789:3-178\)
\(=256+263-178\)
\(=341\)
\(b,1485+\left(735-47\cdot8\right)\)
\(=1485+\left(735-376\right)\)
\(=1485+359\)
\(=1844\)
\(c,5635:5-8172:9\)
\(=1127-908\)
\(=219\)
`A=sqrt{8+2sqrt7}-sqrt{8-2sqrt7}`
`=sqrt{7+2sqrt7+1}-sqrt{7-2sqrt7+1}`
`=sqrt{(sqrt7+1)^2}-sqrt{(sqrt7-1)^2}`
`=sqrt7+1-sqrt7+1=2`
`B=sqrt{11-6sqrt2}+sqrt{6-4sqrt2}`
`=sqrt{9-2.3.sqrt2+2}+sqrt{4-2.2.sqrt2+2}`
`=sqrt{(3-sqrt2)^2}+sqrt{(2-sqrt2)^2}`
`=3-sqrt2+2-sqrt2=5-2sqrt2`
a) 78 060 : (10 – 7) + 300 045
= 78 060 : 3 + 300 045
= 26 020 + 300 045
= 326 065
b) 26 000 + 9 015 × 6
= 26 000 + 54 090
= 80 090