Tính rồi rút gọn
a, 1 + 3 6 =
b, 18 4 - 2 =
c, 3 4 × 4 6 =
d, 3 5 : 4 6 =
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CM
Cao Minh Tâm
10 tháng 5 2019
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Những câu hỏi liên quan
18 tháng 2 2022
1.
a, Ta có 28=14⇒14+56=1312(Bn tự quy đồng lên nhé !)
b,Ta có1216=34⇒18+34=78(Bn tự quy đồng lên nhé !)
c,Ta có636=16⇒14+16=512(Bn tự quy đồng lên nhé !)
d,Ta có1218=23⇒23+14=1112(Bn tự quy đồng lên nhé !)
7.Phần này mk chỉ cho đáp án thuia.67 ;
b,420=15 ;
c,615 ;
d,129
15 tháng 11 2021
\(a,=\dfrac{x^4\left(x-2\right)+2x^2\left(x-2\right)-3\left(x-2\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4+2x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4-x^2+3x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x-1\right)\left(x^2+3\right)}{x+4}\)
\(b,=\dfrac{x^4-3x^2-x^2+3}{x^4-x^2+7x^2-7}=\dfrac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\dfrac{x^2-3}{x^2+7}\\ c,=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}=\dfrac{x^2-1}{x^2+1}\)
25 tháng 7 2018
12/18 + 12/42 = 2/3 + 2/7 = 14/21 + 6/21 = 20/21
1/2 + 2/4 + 3/6 + 4/8 + 5/10 + 6/12
= 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2
= 1/2 x 6
=6/2
=3
NL
1
8 tháng 12 2023
Câu 1:
a: \(\dfrac{2}{5}\sqrt{75}-0,5\cdot\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\cdot\sqrt{12}\)
\(=\dfrac{2}{5}\cdot5\sqrt{3}-0,5\cdot4\sqrt{3}+10\sqrt{3}-\dfrac{2}{3}\cdot2\sqrt{3}\)
\(=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}\)
\(=10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)
b: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
\(=\dfrac{\sqrt{3}\cdot3\sqrt{3}-2\sqrt{3}}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{9-6}\)
\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+3-\sqrt{6}\)
\(=\dfrac{\sqrt{3}}{\sqrt{2}}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)
c: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
=\(\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
Bài 2:
a: 
b: Phương trình hoành độ giao điểm là:
\(3x+2=-x-4\)
=>4x=-6
=>x=-3/2
Thay x=-3/2 vào y=-x-4, ta được:
\(y=-\left(-\dfrac{3}{2}\right)-4=\dfrac{3}{2}-4=-\dfrac{5}{2}\)
Vậy: \(A\left(-\dfrac{3}{2};-\dfrac{5}{2}\right)\)
c: Vì (d2)//(d) nên \(\left\{{}\begin{matrix}a=-1\\b\ne-4\end{matrix}\right.\)
Vậy: (d2): y=-x+b
Thay x=-2 và y=5 vào (d2), ta được:
\(b-\left(-2\right)=5\)
=>b+2=5
=>b=5-2=3
Vậy: (d2): y=-x+3
BT
1
14 tháng 3 2022
24/20+3/20=27/20 khong rut gon duoc nha
6/4-4/5=30/20-16/20=14/20=7/10
2/3x3/4=6/12=1/2
4/10:5/6=4/10x6/5=24/50 khong rut gon duoc
k minh nha
DD
1
YN
15 tháng 2 2023
a) \(\dfrac{5}{6}-\dfrac{2}{14}\)
\(=\dfrac{5}{6}-\dfrac{1}{7}\)
\(=\dfrac{35}{42}-\dfrac{6}{42}\)
\(=\dfrac{29}{42}\)
b) \(\dfrac{5}{20}-\dfrac{1}{6}\)
\(=\dfrac{1}{4}-\dfrac{1}{6}\)
\(=\dfrac{3}{12}-\dfrac{2}{12}\)
\(=\dfrac{1}{12}\)
c) \(\dfrac{5}{9}-\dfrac{3}{12}\)
\(=\dfrac{5}{9}-\dfrac{1}{4}\)
\(=\dfrac{20}{36}-\dfrac{9}{36}\)
\(=\dfrac{11}{36}\)
d) \(8-\dfrac{4}{6}\)
\(=\dfrac{48}{6}-\dfrac{4}{6}\)
\(=\dfrac{44}{6}=\dfrac{22}{3}\).
AT
4 tháng 7 2021
a) \(A=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+-\sqrt{3}-1=-2\)
b) \(B=\sqrt{11-6\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{3^2-2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)
\(=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\left|3-\sqrt{2}\right|-\left|\sqrt{2}-1\right|\)
\(=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)
c) \(C=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left(\sqrt{5}+\sqrt{3}\right)\left|\sqrt{5}-\sqrt{2}\right|\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{2}\right)=5-\sqrt{10}+\sqrt{15}-\sqrt{6}\)
21 tháng 7 2023
a) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
\(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{3^2-\left(\sqrt{5}\right)^2}}\)
\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)
\(=\dfrac{3-\sqrt{5}}{2}\)
b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{2^2-\left(\sqrt{3}\right)^2}}\)
\(=\dfrac{\left|2-\sqrt{3}\right|}{\sqrt{4-3}}\)
\(=\dfrac{2-\sqrt{3}}{1}\)
\(=2-\sqrt{3}\)
21 tháng 7 2023
a: \(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}}=\dfrac{3-\sqrt{5}}{2}\)
b: \(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{1}}=2-\sqrt{3}\)
d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
=(căn 6-11)(căn 6+11)
=6-121=-115
HH
4 tháng 5 2020
Nay rảnh lại qua làm mấy bài lớp 4 ^^
Tui làm luôn kh ghi đề nhá
1,
\(\frac{2}{3}+\frac{8}{15}=\frac{6}{5}\)
\(\frac{2}{5}+\frac{2}{3}=\frac{16}{15}\)
\(\frac{2}{3}+\frac{5}{6}=\frac{3}{2}\)
2,
\(\cdot\frac{3}{7}+\frac{6}{5}+\frac{4}{7}\)
\(=\frac{3}{7}+\frac{4}{7}+\frac{6}{5}\)
\(=1+\frac{6}{5}\)
\(=\frac{11}{5}\)
\(\cdot\frac{6}{4}+\frac{7}{3}+\frac{4}{4}+\frac{3}{3}\)
\(=\frac{6}{4}+\frac{4}{4}+\frac{7}{3}+\frac{3}{3}\)
\(=\frac{10}{4}+\frac{10}{3}\)
\(=\frac{35}{6}\)
8 tháng 1 2022
23+815=6523+815=65
25+23=161525+23=1615
23+56=3223+56=32
2,
⋅37+65+47⋅37+65+47
=37+47+65=37+47+65
=1+65=1+65
=115=115
⋅64+73+44+33⋅64+73+44+33
=64+44+73+33=64+44+73+33
=104+103=104+103
=356
