\(T\text{ìm}\) \(s\text{ố}.nguy\text{ê}n.d\text{ư}\text{ơ}ng.nh\text{ỏ}.nh\text{ất}.th\text{ỏa}.m\text{ãn}:\frac{1}{2}s\text{ố}.\text{đ}\text{ó}.l\text{à}.s\text{ố}.ch\text{ính}.ph\text{ư}\text{ơ}ng\) \(\frac{1}{3}s\text{ố}.\text{đ}\text{ó}.l\text{à}.l\text{ập}.ph\text{ư}\text{ơ}ng.c\text{ủa}.1.s\text{ố}.nguy\text{ên}\) \(\)
\(\frac{1}{5}s\text{ố}.\text{đ}\text{ó}.l\text{à}.l\text{ũy}.th\text{ừa}.5.c\text{ủa}.1.s\text{ố.nguy\text{ê}n}\)
Câu 1: Chứng minh:
\(31.82+125.48+21.43=125.67=1500\)
Câu 2: So sánh:
1,\(\sqrt{51}-\sqrt{5}v\text{à}\sqrt{20}-\sqrt{6}\)
2,\(\sqrt{2}+\sqrt{8}v\text{à}\sqrt{3}+3\)
3,\(\sqrt{37}-\sqrt{14}v\text{à}6-\sqrt{15}\)
4,\(\sqrt{5}+\sqrt{10}v\text{à}5,3\)
So Sánh Các Biểu Thức Sau:
a,\(\sqrt{2}+\sqrt{11}v\text{à}\sqrt{3}+4\) 4
b, \(\sqrt{21}-\sqrt{5}v\text{à}\)\(\sqrt{20}-\sqrt{6}\)
c,\(\sqrt{24}-1v\text{à}\)\(5\)
\(a,\sqrt{2}+\sqrt{11}< \sqrt{3}+\sqrt{16}=\sqrt{3}+4\)
So sánh :
\(a,2^{30}v\text{à}3^{20}\)
\(b,5^{300}v\text{à}3^{500}\)
\(c,2^{24}v\text{à}3^{16}\)
\(d,\left(0,3\right)^{40}v\text{à}\left(0,1\right)^{20}\)
\(\text{a, }2^{30}=8^{10}\)
\(\text{ }3^{20}=\left(3^2\right)^{10}=9^{10}\)
\(\text{Vậy }2^{30}< 3^{20}\)
\(\text{b, }5^{300}=\left(5^3\right)^{100}=125^{100}\)
\(3^{500}=\left(3^5\right)^{100}=243^{100}\)
\(\text{Vậy }5^{300}< 243^{100}\)
\(\text{c, }2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
\(\text{Vậy ...}\)
Không dùng máy tính ,hãy so sánh :
1 )\(\sqrt{7-\sqrt{21}+4\sqrt{5}}v\text{à}\sqrt{5}-1\)
2 )\(\sqrt{5}+\sqrt{10}+1v\text{à}\sqrt{35}.\)
3 )\(\frac{15-2\sqrt{10}}{3}v\text{à}\sqrt{15}.\)
1) \(A=\left(\sqrt{7-\sqrt{21}+4\sqrt{5}}\right)^2=7-\sqrt{21}+4\sqrt{5}\)
\(B=\left(\sqrt{5}-1\right)^2=6-2\sqrt{5}\)
\(\Rightarrow A-B=1-\sqrt{21}+6\sqrt{5}=\left(1+\sqrt{180}\right)-\sqrt{21}>0\)
\(\Rightarrow A>B\Rightarrow\sqrt{7-\sqrt{21}+4\sqrt{5}}>\sqrt{5}-1\)
2) \(C=\left(\sqrt{5}+\sqrt{10}+1\right)^2=5+10+1+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}\)
\(=26+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}>26+10>35=\left(\sqrt{35}\right)^2\)
Vậy \(\sqrt{5}+\sqrt{10}+1>\sqrt{35}\)
3) \(\left(\frac{15-2\sqrt{10}}{3}\right)^2=\frac{225-60\sqrt{10}+40}{9}=\frac{265-60\sqrt{10}}{9}=\frac{265}{9}-\frac{20\sqrt{10}}{3}< 15\)
Vậy nên \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)
a)\(\frac{z}{5}=\frac{x}{2}=\frac{y}{3}v\text{à}x.y-z=810\)
b)\(5x=3yv\text{à}2x^2-y^2=-28\)
c)\(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}v\text{à}x^2+y^2+z^2=14\)
d)\(x:y:z=3:4:5v\text{à}5z^2-2y^2=594\)
Tìm x, y, z
a, \(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}v\text{à}2\text{x}+3y-z=186\)
b, 3x=2y ; 7y = 5z và x-y+z = 32
c,\(\frac{2\text{x}}{3}=\frac{3y}{4}=\frac{4\text{z}}{5}v\text{à}x+y+z=49\)
d, \(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}v\text{à}x^2+y^2+z^2=14\)
e, x+y=x:y= 3.(x-y)
b, \(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\)
\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)
\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
áp dụng dãy tỉ số bằng nhau :
\(\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)
x = 2 . 10 = 20
y = 2 . 15 = 30
z = 2 . 21 = 42
Vậy : .....
a, \(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)
MSC của y là : 20
Có: \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
Áp dụng dãy tỉ số bằng nhau, ta có:
\(2x+3y-z=186\)
\(\Rightarrow2.15+3.20-28=30+60-28=62\)
\(\frac{186}{62}=3\)
x = 3 . 15 = 45
y = 3 . 20 = 60
z = 3 . 28 = 84
Vậy: .....
So sánh :
\(10^{30}v\text{à}2^{100}\)
\(5^{300}v\text{à}3^{453}\)
\(29^{12}v\text{à}18^{17}\)
103và 2100
Ta có:1030=(103)10=100010
2100=(210)10=102410
Vì 1000<1024 nên 1030<2100
5300 và 3453
Ta có:5300=(52)150=25150
3453=(33)151=27151=27.27150
Vì 25 < 27.27 nên 5300<3453
nhớ k ch mình nhé
so sánh
a, \(\sqrt{2}+\sqrt{3}+\sqrt{5}v\text{à}18\)
b, \(\sqrt{5}+\sqrt{7}+4v\text{à}12\)
\(\sqrt{2}+\sqrt{3}+\sqrt{5}< \sqrt{4}+\sqrt{9}+\sqrt{25}=2+3+5=10< 18\)
b) \(\sqrt{5}+\sqrt{7}+4< \sqrt{9}+\sqrt{9}+4=3+3+4=10< 12\)
Vì \(\sqrt{2}\) và các căn bậc khác đều là nhưng số thực nên ta cha nó là
\(\sqrt{2}+\sqrt{3}+\sqrt{5}=2+3+5=10\)
Mà 10<12
\(\Rightarrow dpcm\)
