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a) \(9=6+3=6+\sqrt{9}\)
\(6+2\sqrt{2}=6+\sqrt{8}\)
\(\sqrt{8}< \sqrt{9}\) nên \(6+\sqrt{8}=6+2\sqrt{2}< 6+\sqrt{9}=9\)
b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)
\(3^2=9=5+4=5+\sqrt{16}\)
\(\sqrt{16}< \sqrt{24}\Rightarrow3^2< \left(\sqrt{2}+\sqrt{3}\right)^2\Rightarrow3< \sqrt{2}+\sqrt{3}\)
c) \(9+4\sqrt{5}=\left(2+\sqrt{5}\right)^2\)
\(16=\left(2+2\right)^2=\left(2+\sqrt{4}\right)^2\)
\(\sqrt{4}< \sqrt{5}\Rightarrow2+\sqrt{4}< 2+\sqrt{5}\Rightarrow\left(2+\sqrt{4}\right)^2=16< \left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)
d) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)
\(2^2=14-10=14-\sqrt{100}\)
\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)
\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)
1.a)
\(2\sqrt{3}=\sqrt{12}>\sqrt{9}=3.\)
\(3\sqrt{2}=\sqrt{18}>\sqrt{16}=4.\)
Suy ra VT > 7
1.b)
\(\sqrt{16}+\sqrt{25}=4+5=9\)
2.a)
\(\sqrt{21-6\sqrt{6}}=\sqrt{\left(3\sqrt{2}\right)^2-6\sqrt{6}+3}=3\sqrt{2}-\sqrt{3}\)
b)\(\sqrt{9-2\sqrt{14}}=\sqrt{\frac{18-4\sqrt{14}}{2}}=\frac{\sqrt{14}-2}{\sqrt{2}}=\sqrt{7}-1\)
Các câu còn lại bạn làm tương tự nhé!
c) \(\sqrt{4-\sqrt{7}}=\frac{1}{\sqrt{2}}.\sqrt{8-2\sqrt{7}}=\frac{1}{\sqrt{2}}\sqrt{7-2\sqrt{7}+1}\)
\(=\frac{1}{\sqrt{2}}\sqrt{\left(\sqrt{7}-1\right)^2}=\frac{\sqrt{2}\left(\sqrt{7}-1\right)}{2}\)
d) \(\sqrt{4+2\sqrt{3}-\sqrt{4-2\sqrt{3}}}=\sqrt{4+2\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{4+2\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{4+2\sqrt{3}-\sqrt{3}+1}=\sqrt{5+\sqrt{3}}\)
A = ((20 + 1) . 20 : 2) . 2 = 420
B = (25 + 20) . 6 : 2 = 135
C = ( 33 + 26) . 8 : 2 = 236
D = (1 + 100) .100 : 2 = 5050
a/ \(\left(x-2\right)^2=11+6\sqrt{2}\)
\(\Leftrightarrow\left(x-2\right)^2=\left(3+\sqrt{2}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3+\sqrt{2}\\x-2=-3-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5+\sqrt{2}\\x=-1-\sqrt{2}\end{matrix}\right.\)
b/ \(x^2-10x+25=27-10\sqrt{2}\)
\(\Leftrightarrow\left(x-5\right)^2=\left(5-\sqrt{2}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=5-\sqrt{2}\\x-5=\sqrt{2}-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10-\sqrt{2}\\x=\sqrt{2}\end{matrix}\right.\)
c/ \(4x^2+4x+1=28-10\sqrt{3}\)
\(\Leftrightarrow\left(2x+1\right)^2=\left(5-\sqrt{3}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=5-\sqrt{3}\\2x+1=\sqrt{3}-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4-\sqrt{3}}{2}\\x=\frac{-6+\sqrt{3}}{2}\end{matrix}\right.\)
d/ \(x^2+2\sqrt{5}x+5=21-4\sqrt{5}\)
\(\Leftrightarrow\left(x+\sqrt{5}\right)^2=\left(2\sqrt{5}-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{5}=2\sqrt{5}-1\\x+\sqrt{5}=1-2\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}-1\\x=1-3\sqrt{5}\end{matrix}\right.\)
e/ \(x^2+2\sqrt{12}x+12=13-4\sqrt{3}\)
\(\Leftrightarrow\left(x+2\sqrt{3}\right)^2=\left(2\sqrt{3}-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2\sqrt{3}=2\sqrt{3}-1\\x+2\sqrt{3}=1-2\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1-4\sqrt{3}\end{matrix}\right.\)
f/ \(4x^2-12\sqrt{2}x+18=51-10\sqrt{2}\)
\(\Leftrightarrow\left(2x-3\sqrt{2}\right)^2=\left(5\sqrt{2}-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5\sqrt{2}=5\sqrt{2}-1\\2x-2\sqrt{2}=1-5\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{10\sqrt{2}-1}{2}\\x=\frac{1-3\sqrt{2}}{2}\end{matrix}\right.\)
Đặt A(x)= P(x) - x2= 0
Có: A(1)=P(1) -12 =0
A(2) = P(2) -22=0
A(3)=P(3)-32=0
A(4)=P(4)-44=0
A(5)=P(5)-55=0
=> x thuộc {1;2;3;4;5} là nghiệm của A(x)
=> A(x)=(x-1)(x-2)(x-3)(x-4)(x-5)=P(x)-x2
P(x)= (x-1)(x-2)(x-3)(x-4)(x-5)+x2
P(6)=156
P(7)=769
P(8)=2584
P(9)=6801
a/ \(A=\frac{30\left(\sqrt{6}-1\right)}{5}+\frac{2\left(\sqrt{6}+2\right)}{2}-\frac{6\left(3+\sqrt{6}\right)}{3}=6\sqrt{6}-6+\sqrt{6}+2-6-2\sqrt{6}\)
\(A=5\sqrt{6}-10\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{\left(2\sqrt{2}+1\right)^2}}=\sqrt{18-4\sqrt{2}}\)
Đến đây ko rút gọn được nữa, nhưng nếu đề là:
\(B=\sqrt{17+6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}=\sqrt{18+8\sqrt{2}}=4+\sqrt{2}\)
c/
\(C=\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(C=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\)
\(D=\sqrt{a-2\sqrt{a}+1}-\sqrt{a-8\sqrt{a}+16}\)
\(D=\sqrt{\left(\sqrt{a}-1\right)^2}-\sqrt{\left(4-\sqrt{a}\right)^2}=\sqrt{a}-1-\left(4-\sqrt{a}\right)=2\sqrt{a}-5\)
\(E=\sqrt{a-2+2\sqrt{a-2}+1}+\sqrt{a-2-2\sqrt{a-2}+1}\) (\(a\ge2\))
\(E=\sqrt{\left(\sqrt{a-2}+1\right)^2}+\sqrt{\left(\sqrt{a-2}-1\right)^2}\)
\(E=\sqrt{a-2}+1+\left|\sqrt{a-2}-1\right|\)
\(\Rightarrow\left[{}\begin{matrix}E=2\sqrt{a-2}\left(a\ge3\right)\\E=2\left(2\le a\le3\right)\end{matrix}\right.\)
\(F=\sqrt[3]{10+6\sqrt{3}}-\sqrt{3}=\sqrt[3]{1+3.1.\sqrt{3}+3.1.\sqrt{3}^2+\sqrt{3}^3}-\sqrt{3}\)
\(F=\sqrt[3]{\left(1+\sqrt{3}\right)^3}-\sqrt{3}=1+\sqrt{3}-\sqrt{3}=1\)
\(G=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\Rightarrow G^3=\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)^3\)
\(\Rightarrow G^3=14+3\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\left(\sqrt[3]{49-50}\right)\)
\(\Rightarrow G^3=14-3G\Rightarrow G^3+3G-14=0\)
\(\Rightarrow G=2\)
a) \(\left\{{}\begin{matrix}3x-4y=-2\\2x+y=6\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}3x-4y=-2\\8x+4y=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=22\\3x-4y=-2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
a: =>3x-4y=-2 và 8x+4y=24
=>11x=22 và 2x+y=6
=>x=2 và y=6-2x=6-2*2=2
b: 2x-y=0 và 3x+y=4
=>5x=4 và y=2x
=>x=4/5 và y=8/5
c: x+3y=-2 và x-y=-1
=>4y=-1 và x=y-1
=>y=-1/4 và x=-1/4-1=-5/4
d: x+y=3 và 4x-3y=-2
=>4x+4y=12 và 4x-3y=-2
=>7y=14 và x+y=3
=>y=2 và x=1
Tính :\(a,\)\(-\sqrt{\left(-6\right)^2}=-|-6|=-6\)
\(b,\)\(-\sqrt{\frac{-25}{-16}}=-\sqrt{\left(\frac{5}{4}\right)^2}=-|\frac{5}{4}|=-\frac{5}{4}\)
\(c,\)\(\sqrt{-\frac{-9}{25}}=\sqrt{\frac{9}{25}}=\sqrt{\left(\frac{3}{5}\right)^2}=|\frac{3}{5}|=\frac{3}{5}\)
\(d,\)\(\left(-\sqrt{7}\right)^2=7\)
\(e,\)\(-\left(\frac{\sqrt{3}}{4}\right)^2=-\frac{\sqrt{3}^2}{4^2}=-\frac{3}{16}\)
\(f,\)\(\sqrt{\left(-2\right)^4}=\sqrt{\left[\left(-2\right)^2\right]^2}=|-2^2|=4\)
So sánh :\(a,\) \(\sqrt{8}-1\)
\(2=3-1=\sqrt{9}-1\)
\(\Rightarrow\sqrt{8}-1< 2\)
\(b,\)\(\sqrt{\frac{16}{2}}=\sqrt{8}>\sqrt{3}\)
\(\Rightarrow\sqrt{\frac{16}{2}}>\sqrt{3}\)