1 ) tính
a ) 62 - 4 x 6 + 48
b ) 96 - 32 : 4 - 49
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a) 64 - 4 x 6 - 48
= 4 x 16 - 4 x 6 - 48
= 4 x (16 - 6) - 48
= 4 x 10 - 48
= 40 - 48
= -8
b) 96 - 32 : 4 - 49
= 96 - 8 - 49
= 39
Tính
a )
64 - 4 x 6 - 48
= 64 - 24 - 48
= 40 - 48
= -8
b )
96 - 32 : 4 - 49
= 96 - 8 - 49
= 88- 49
=39
tk mk nha
mơn nhiều ạ
\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}=6\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)
\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}=2\sqrt{5}+4\sqrt{2}\)
a: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
b: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}\)
=6
c: Ta có: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}\)
=10
d: Ta có: \(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+4\sqrt{90}}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}\)
\(=2\sqrt{5}+4\sqrt{2}\)
a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )
Vậy x = 1
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)
=> x + 100 = 0
=> x = -100
c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)
=> x - 100 = 0
=> x = 100
Chúc bạn học tốt
có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai
a)16.4x=48
\(4^2.4^x=4^{8^{ }}\)
\(4^x=4^{8^{ }}:4^2\)
\(4^x=4^{6^{ }}\)
\(x=6\)
b)(X-2)(X-5)=0
\(\Rightarrow x-2=0\rightarrow x=2\)
\(x-5=0\rightarrow x=5\)
Vậy x∈ {2;5}
c)2x+2x+1=96
\(2^x.1+2^{x^{ }}.2\)
\(2^x.\left(1+2\right)=96\)
\(2^x.3=96\)
\(2^x=96:3\)
\(2^x=32\)
\(2^x=2^5\)
\(x=5\)
\(Zzz\) 😪
chị giúp em đc mấy bài trong trang của em mới đăng đc ko ạ?
Tính hợp lí
a,15.(27+ 18 +6 ) + 15.( 23 +12)
= 15(27+18+6+23+12)
= 15.86
= 1290
b,24.(15 + 49 ) + 12. ( 50 + 42 )
=12.2(15+49)+12.(50+42)
= 12(30+98)+12(50+42)
=12(30+98+50+42)
=12.220
= 2640
c,53.(51+4)+53.(49+96) +53
=53(51+4)+53(49+96)+53.1
=53(51+4+49+96+1)
=53.201
=10653
d,42.( 15 +96) +6.(25 +4).7
=42(15+96)+42(25+4)
=42(15+96+25+4)
=42.140
=5880
tick đúng cho mk nha pikachu
Câu 4:
\(4^{n+2}-4^{n-1}=252\\ \Leftrightarrow4^{n-1}.\left(4^3-1\right)=252\\ \Leftrightarrow4^{n-1}.63=252\\ \Leftrightarrow4^{n-1}=\dfrac{252}{63}=4=4^1\\ \Rightarrow n-1=1\\ \Rightarrow n=2\)
a ) 62 - 4 x 6 + 48 b ) 96 - 32 : 4 - 49
= 62 - 24 + 48 = 96 - 8 - 49
= 38 + 48 = 86 - 49
= 86 = 37
ai thấy đúng thì tk mk nha
ai tk mk mk tk lại
a ) 86
b ) 39