(√12- 2√3+5√2 -3/4√8)2√6
Rút gọn biểu thức
Help me 😭
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=2\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=4\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}\)
b: \(=2\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}\)
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) \(8-\dfrac{4}{6}\)
\(=\dfrac{48}{6}-\dfrac{4}{6}\)
\(=\dfrac{44}{6}=\dfrac{22}{3}\).
\(a,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\\ =\sqrt{3}+\sqrt{2}-\left(\sqrt{3}-\sqrt{2}\right)\\ =\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\\=2\sqrt{2} \)
\(b,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)
\(c,=x-4+\sqrt{\left(4^2-2.4.x+x^2\right)}\\ =x-4+\sqrt{\left(4-x\right)^2}\\ =x-4+\left|4-x\right|\\ =x-4+x-4=2x-8\) (vì \(x>4\) )
@seven
P = 8.( 7 - 72 + 73 - 74 +...+ 72022)
Đặt B = 7 - 72 + 73 - 74+...+ 72022
7 \(\times\)B = 72 - 73 + 74-....- 72022 + 72023
7B + B = 7 + 72023
8B = ( 7 + 72023)
B = ( 7 + 72023): 8
P = 8 \(\times\) ( 7 + 72023) : 8
P = 7 + 72023
-0,26
p/s: bấm máy tính có thể sai chỗ nào đó nên sai không chịu trách nhiệm đâu nhé @@
\(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
=(căn 6-11)(căn 6+11)
=6-121=-115
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}\right)^2-1^2}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}\right)^2-2^2}-\dfrac{12\left(3+\sqrt{6}\right)}{3^2-\left(\sqrt{6}\right)^2}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}\right)^2-11^2\)
\(=6-121\)
\(=-115\)
\(\frac{2^{12}\cdot3^5-\left(2^2\right)^6.3^5.3}{2^{12}.\left(3^2\right)^3+\left(2^3\right)^4.3^5}\)
=\(\frac{2^{12}\cdot3^5-2^{12}.3^5.3}{2^{12}.3^5+2^{12}.3^5}\)
=3
\(=\left(2\sqrt{3}-2\sqrt{3}+5\sqrt{2}-\frac{3}{4\sqrt{8}}\right)2\sqrt{6}\)
=\(5\sqrt{2}.2\sqrt{6}-\frac{3}{8\sqrt{2}}.2\sqrt{2}.\sqrt{3}\)
=\(20\sqrt{3}-\frac{3\sqrt{3}}{4}=\sqrt[]{3}.\left(20-\frac{3}{4}\right)=\frac{\sqrt{3}.77}{4}\)