0,1.√225-√1/4
Tính giá trị của biểu thức:
a , 225 × 4 − 378 : 6 + 165 b , 48 , 7 + 25 , 8 × 0 , 1 + 0 , 63
a , 225 × 4 − 378 : 6 + 165 = 225 × 4 − 63 + 165 = 225 × 4 − 228 = 900 − 228 = 672 b , 48 , 7 + 25 , 8 × 0 , 1 + 0 , 63 = 74 , 5 × 0 , 1 + 0 , 63 = 7 , 45 + 0 , 63 = 8 , 08
Tính giá trị của biểu thức:
a) 225 × 4 − 378 : 6 + 165
b) 48 , 7 + 25 , 8 × 0 , 1 + 0 , 63
a) 225 × 4 − 378 : 6 + 165 = 225 × 4 − 63 + 165
= 225 × 4 − 228
= 900 − 228 = 672
b) 48 , 7 + 25 , 8 × 0 , 1 + 0 , 63 = 74 , 5 × 0 , 1 + 0 , 63
= 7 , 45 + 0 , 63
= 8 , 08
1.thực hiện phép tính
a.(0,125).(-3.7).(-2)\(^3\)
b.\(\sqrt{36}.\sqrt{\frac{25}{16}}+\frac{1}{4}\)
c.\(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1^2_5\)
d.\(0,1.\sqrt{225}.\sqrt{\frac{1}{4}}\)
Bài 1:
a) Ta có: \(\left(0.125\right)\cdot\left(-3\cdot7\right)\cdot\left(-2\right)^3\)
\(=\frac{1}{8}\cdot\left(-21\right)\cdot\left(-8\right)\)
\(=\frac{1}{8}\cdot168\)
\(=21\)
b) Ta có: \(\sqrt{36}\cdot\sqrt{\frac{25}{16}}+\frac{1}{4}\)
\(=\sqrt{36\cdot\frac{25}{16}}+\frac{1}{4}\)
\(=\sqrt{\frac{225}{4}}+\frac{1}{4}\)
\(=\frac{15}{2}+\frac{1}{4}\)
\(=\frac{31}{4}\)
c) Ta có: \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}=-1\)
d) Ta có: \(0,1\cdot\sqrt{225}\cdot\sqrt{\frac{1}{4}}\)
\(=0,1\cdot15\cdot\frac{1}{2}=\frac{3}{4}\)
TÍNH GIÁ TRỊ CÁC BIỂU THỨC SAU
A,\(\sqrt{0,09}-\sqrt{0,64}\)
B,\(0,1\times\sqrt{225}-\sqrt{\frac{1}{4}}\)
C,\(\sqrt{0,36}\times\sqrt{\frac{25}{16}+\frac{1}{4}}\)
D,\(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}-1\frac{2}{5}}\)
a) \(\sqrt{0,09}-\sqrt{0,64}=\frac{-1}{2}=-0,5\)
b) \(0,1\cdot\sqrt{225}-\sqrt{\frac{1}{4}}=0,1\cdot15-\frac{1}{2}=1\)
c) \(\sqrt{0,36}\cdot\sqrt{\frac{25}{16}+\frac{1}{4}}=\frac{3\sqrt{29}}{20}\)
d) đề baì có sai ko ban?
a) A = \(3\frac{1}{117}.4\frac{1}{119}-1\frac{116}{117}.5\frac{118}{119}-\frac{5}{119}\)
b) B = \(4\frac{1}{115}.1\frac{1}{225}-5\frac{114}{115}.1\frac{224}{225}-\frac{10}{225}\)
Tính :\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+....+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
Sorry mới lớp 6 chưa học
thông cảm
no chửi
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào bài toán ta được
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)
\(=1-\frac{1}{\sqrt{225}}=1-\frac{1}{15}=\frac{14}{15}\)
Tính giá trị biểu thức :
a ) A = \(3\frac{1}{7}.4\frac{1}{119}-1\frac{116}{117}.5\frac{118}{119}-\frac{5}{119}\)
b ) B = \(4\frac{1}{115}.3\frac{1}{225}-5\frac{114}{115}.1\frac{224}{225}-\frac{10}{225}\)
Tính giá trị biểu thức sau :
a) A = \(3\dfrac{1}{117}.4\dfrac{1}{119}-1\dfrac{116}{117}.5\dfrac{118}{119}-\dfrac{5}{119}\)
b) B = \(4\dfrac{1}{115}.3\dfrac{1}{225}-5\dfrac{114}{115}.1\dfrac{224}{225}-\dfrac{10}{225}\)
(1/6+0,1+1/15):(1,6+0,1-1/15)/(0,5-1/3+1/4-1/5):(1/4-1/6)
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