a, Thực hiện phép tính A = 7 - 4 3 + 1 2 - 3
b, Rút gọn biểu thức B = sin 2 19 0 + cos 2 19 0 + tan 19 0 - c o t 71 0
K
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H9
HT.Phong (9A5)
CTVHS
5 tháng 9 2023
a) \(\sqrt{2}\left(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\right)\)
\(=\sqrt{2\cdot\left(4+\sqrt{7}\right)}+\sqrt{2\cdot\left(4-\sqrt{7}\right)}\)
\(=\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2+2\cdot\sqrt{7}\cdot1+1^2}+\sqrt{\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot1+1^2}\)
\(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=\left|\sqrt{7}+1\right|+\left|\sqrt{7}-1\right|\)
\(=\sqrt{7}+1+\sqrt{7}-1\)
\(=2\sqrt{7}\)
b) \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2\cdot\left(2-\sqrt{3}\right)}-\sqrt{2\cdot\left(2+\sqrt{3}\right)}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{ }\)
\(=-\dfrac{2}{\sqrt{2}}\)
\(=-\sqrt{2}\)
AH
Akai Haruma
Giáo viên
18 tháng 9 2023
Lời giải:
a.
\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)
$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.
$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$
$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.
$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$
$=1(3+\sqrt{2})=3+\sqrt{2}$
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 10 2023
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 10 2023
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
6 tháng 9 2023
a: \(=\dfrac{2\sqrt{2}+3+2\sqrt{2}-3}{8-9}\)
\(=\dfrac{4\sqrt{2}}{-1}=-4\sqrt{2}\)
b: \(=\dfrac{\sqrt{2}\left(2\sqrt{2}-\sqrt{7}\right)+\sqrt{2}\left(2\sqrt{2}+\sqrt{7}\right)}{8-7}\)
\(=4-\sqrt{14}+4+\sqrt{14}=8\)
c: \(=\dfrac{2+\sqrt{5}-2\left(2-\sqrt{5}\right)}{-1}=\dfrac{2+\sqrt{5}-4+2\sqrt{5}}{-1}\)
\(=-3\sqrt{5}+2\)
27 tháng 8 2023
a: \(=\left(\sqrt{3}-2\right)\cdot\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\)
=3-4=-1
b: \(=\sqrt{6+4\sqrt{2}}-\sqrt{11-2\sqrt{18}}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=2+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}-1\)
c: \(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
H
27 tháng 10 2023
Bài `1`
\(\sqrt{4-2\sqrt{3}}-\dfrac{2}{\sqrt{3}+1}+\dfrac{\sqrt{3}-3}{\sqrt{3}-1}\\ =\sqrt{3-2\sqrt{3}+1}-\dfrac{2\left(\sqrt{3}-1\right)}{3-1}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}+1-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}+1-\sqrt{3}\\ =-\sqrt{3}\)
27 tháng 10 2023
2:
a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)
b: B=5
=>\(5\left(\sqrt{x}+3\right)=\sqrt{x}+8\)
=>\(5\sqrt{x}+15=\sqrt{x}+8\)
=>\(4\sqrt{x}=-7\)(loại)
Vậy: \(x\in\varnothing\)
10 tháng 8 2023
26:
A=12x^2+10x-6x-5-(12x^2-8x+3x-2)
=12x^2+4x-5-12x^2+5x+2
=9x-3
Khi x=-2 thì A=-18-3=-21
25:
b: \(\left(y-3\right)\left(y^2+y+1\right)-y\left(y^2-2\right)\)
=y^3+y^2+y-3y^2-3y-3-y^3+2y
=-2y^2-3
với x ³ 0 và x ¹ 1.
với x > 2017
a, A = 7 - 4 3 + 1 2 - 3 = 2 - 3 + 2 + 3 = 4
b, B = sin 2 19 0 + cos 2 19 0 + tan 19 0 - c o t 71 0
= sin 2 19 0 + cos 2 19 0 + tan 19 0 - tan 19 0 = 1