Tính giá trị của biểu thức:
a) 31 + 24 x 2 =
b) (90 - 50) : 5 =
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a) 4 500 : 90 : 25 = 50 : 25 = 2
b) 840 : (3 x 4) = 840 : 12 = 70
c) 682 + 96 : 12 = 682 + 8 = 690
d) 2 784 : 24 – 16 = 116 – 16 = 100
a) \(36 – 18:6 = 36 – 3 = 33\)
b) \(2.3^{2} + 24 : 6 . 2 = 2.9 + 4.2 = 18 + 8 = 26\)
c) \(2.3^{2} – 24 : (6.2) = 2.9 - 24 : 12 = 18 - 2 = 16.\)
a) 96 : 3 x 5 = 32 x 5 = 160 | b) 60 : (2 x 3) = 60 : 6 = 10 |
b) Thay x=-28
\(\left(-2\right)+17+\left(-28\right)=-13\)
c) Thay x=-4
\(\left(-4\right)+25+\left(-47\right)=-26\)
d) Thay x=8
\(25+8+\left(-13\right)=20\)
`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`
`b)` Với `x ne -1;x ne -5` có:
`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`
`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`
`A=[x^2-3x-4]/[(x+1)(x+5)]`
`A=[(x+1)(x-4)]/[(x+1)(x+5)]`
`A=[x-4]/[x+5]`
`c)` Với `x ne -5; x ne -1; x ne 4` có:
`P=A.B=[x-4]/[x+5].[-10]/[x-4]`
`=[-10]/[x+5]`
Để `P` nguyên `<=>[-10]/[x+5] in ZZ`
`=>x+5 in Ư_{-10}`
Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`
`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)
a) 9 234 : [3 . 3. (1 + 83)] = 9 234 : [3 . 3 . (1 + 512)]
= 9 234 : [3 . 3 . 513] = 9 234 : 4617 = 2
b) 76 - {2 . [2 . 52 - (31 - 2 . 3)]} + 3 . 25
= 76 - {2 . [2 . 25 - (31 - 6)]} + 75
= 76 - {2 . [50 - 25]} + 75 = 76 - {2 . 25} + 75 = 76 - 50 + 75 = 101
a) \(\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)
\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)
\(=\dfrac{9}{5}+1\)
\(=\dfrac{14}{5}\)
b) \(\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)
\(=\dfrac{7}{5}-\dfrac{1}{6}\)
\(=\dfrac{42}{30}-\dfrac{5}{30}\)
\(=\dfrac{37}{30}\)
\(a,\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)
\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)
\(=\dfrac{9}{5}+1\)
\(=\dfrac{9}{5}+\dfrac{5}{5}\)
\(=\dfrac{14}{5}\)
\(b,\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)
\(=\dfrac{7}{5}-\dfrac{1}{6}\)
\(=\dfrac{42}{30}-\dfrac{5}{30}\)
\(=\dfrac{37}{30}\)
\(a,A=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)
\(=4\sqrt{2}\)
\(b,B=\left|1-\sqrt{5}\right|+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|1-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left|1-\sqrt{5}\right|+\left|\sqrt{5}+1\right|=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
\(c,C=\dfrac{2+\sqrt{6}+2-\sqrt{6}}{\left(2+\sqrt{6}\right)\left(2-\sqrt{6}\right)}=\dfrac{4}{4-6}=-2\)
Lời giải:
a.
\(A=2\sqrt{2}-3\sqrt{18}+4\sqrt{32}-\sqrt{50}=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)
\(=(2-9+16-5)\sqrt{2}=4\sqrt{2}\)
b.
\(B=\sqrt{(1-\sqrt{5})^2}+\sqrt{(\sqrt{5}+1)^2}=|1-\sqrt{5}|+|\sqrt{5}+1|=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
c.
\(C=\frac{2+\sqrt{6}+2-\sqrt{6}}{(2-\sqrt{6})(2+\sqrt{6})}=\frac{4}{2^2-6}=-2\)
a) 31 + 24 x 2 = 31 + 48 (0,25đ)
= 79 (0,25đ)
b) (90 - 50) : 5 = 40 : 5 (0,25đ)
= 8 (0,25đ)