Tính và so sánh
a) \(\sqrt{\dfrac{169}{64}}\) và \(\dfrac{\sqrt{169}}{\sqrt{64}}\)
b)\(\sqrt{\dfrac{2,25}{2,56}}\) và \(\dfrac{\sqrt{2,25}}{\sqrt{2.56}}\)
giúp mình với
1, tính
a, \(7\times\sqrt{\dfrac{6^2}{7^2}}-\sqrt{25}+\sqrt{\dfrac{\left(-3\right)^2}{2}}\)
b, \(-\sqrt{\dfrac{64}{49}}-\dfrac{3}{5}\times\sqrt{\dfrac{25}{64}}+\sqrt{0,25}\)
c, \(\sqrt{\dfrac{10000}{5}}-\dfrac{1}{4}.\sqrt{\dfrac{16}{9}}+\sqrt{\dfrac{\left(-3\right)^2}{\left(4\right)}}\)
d, \(\left|\dfrac{1}{4}-\sqrt{0,0144}\right|-\dfrac{3}{2}+\sqrt{\dfrac{81}{169}}\)
a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)
b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)
\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)
c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)
Không dùng mtct, hãy so sánh
A=\(\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\)và 20
B=\(\sqrt{196}-\dfrac{1}{\sqrt{3}}-1\)và c=\(\sqrt{169}+\dfrac{-1}{\sqrt{2}}\)
M=\(\sqrt{61-35}\)vàN=\(\sqrt{61}-\sqrt{35}\)
giúp mình với
1, tính
a, \(7\times\sqrt{\dfrac{6^2}{7^2}}-\sqrt{25}+\sqrt{\dfrac{\left(-3\right)^2}{2}}\)
b, \(-\sqrt{\dfrac{64}{49}}-\dfrac{3}{5}\times\sqrt{\dfrac{25}{64}}+\sqrt{0,25}\)
c, \(\sqrt{\dfrac{10000}{5}}-\dfrac{1}{4}.\sqrt{\dfrac{16}{9}}+\sqrt{\dfrac{\left(-3\right)^2}{\left(4\right)}}\)
d, \(\left|\dfrac{1}{4}-\sqrt{0,0144}\right|-\dfrac{3}{2}+\sqrt{\dfrac{81}{169}}\)
câu 1 . thực hiện phép tính
a. \(\sqrt{4,9.360}\)
b. \(\sqrt{2,25.0,04}\)
c. \(\sqrt{3\dfrac{1}{16}.2\dfrac{14}{15}}\)
d. \(\sqrt{9-\sqrt{17}}-\sqrt{9+\sqrt{17}}\)
e. \(\sqrt{\dfrac{144}{169}}\)
g.\(\dfrac{\sqrt{27}}{\sqrt{3}}\)
f. \(\sqrt{2,25}\)
n.\(\sqrt{\dfrac{25}{529}}\)
\(a,\sqrt{4,9.360}=\sqrt{49.36}=\sqrt{49}.\sqrt{36}=7.6=42\)
b,\(\sqrt{2,25.0,04}=\sqrt{0.09}=0.3\)
c, \(\sqrt{3\dfrac{1}{16}.2\dfrac{4}{15}}=\sqrt{\dfrac{49}{16}.\dfrac{44}{15}}=\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{44}{15}}=\dfrac{7}{4}.1,7=2,99\approx3\)
e, \(\sqrt{\dfrac{144}{169}}=\dfrac{\sqrt{144}}{\sqrt{169}}=\dfrac{12}{13}\)
g,\(\dfrac{\sqrt{27}}{\sqrt{3}}=\sqrt{\dfrac{27}{3}}=\sqrt{9}=3\)
f,\(\sqrt{2,25}=\dfrac{3}{2}\)
n,\(\sqrt{\dfrac{25}{529}}=\dfrac{\sqrt{25}}{\sqrt{529}}=\dfrac{5}{23}\)
bài 10: So sánh
2 \(\dfrac{1}{2}\)và 2,25 ; \(\sqrt{6}\) và \(\sqrt{5}\) ; \(-\sqrt{5}\) và \(\sqrt{5}\) ; 6,2 và \(\sqrt{36,6}\)
bài 11: tìm số thích hợp điền vào dấu ... :
a) -1,2 ... 8 > - 1,247
b) -1,15 < -1, ... 5
c) -3,14 (91) > -3,1491 ...
GIẢI GẤP GIÚP MÌNH Ạ, MÌNH CẦN GẤP
11:
a: Số cần tìm là 3;2;1;0
b: 0
10:
\(2\dfrac{1}{2}=2.5>2.25\)
\(6>5\)
nên căn 6>căn 5
-căn 5<0
0<căn 5
=>-căn 5<căn 5
Với \(x>0\) cho 2 biểu thức \(A=\dfrac{2+\sqrt{x}}{\sqrt{x}}\) và \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
1) Tính giá trị của biểu thức A khi \(x=64\)
2) Rút gọn biểu thức B
3) Tìm x để \(\dfrac{A}{B}>\dfrac{3}{2}\)
1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)
2: \(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
3: A/B>3/2
=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{\sqrt{x}\cdot2}>0\)
=>\(-\sqrt{x}+2>0\)
=>-căn x>-2
=>căn x<2
=>0<x<4
1) Thay x=64 vào A ta có:
\(A=\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)
2) \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
3) Ta có:
\(\dfrac{A}{B}>\dfrac{3}{2}\) khi
\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)
\(\Leftrightarrow\dfrac{2-\sqrt{x}}{2\sqrt{x}}>0\)
Mà: \(2\sqrt{x}\ge0\forall x\)
\(\Leftrightarrow2-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Leftrightarrow x< 4\)
Kết hợp với đk:
\(0< x< 4\)
So sánh A và B biết :
\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)
\(B=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)
Ta có: \(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{120}+11\)
=10
Ta có: \(B=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)
\(=\dfrac{2}{\sqrt{1}+\sqrt{1}}+\dfrac{2}{\sqrt{2}+\sqrt{2}}+...+\dfrac{2}{\sqrt{35}+\sqrt{35}}\)
\(\Leftrightarrow B< 2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{35}+\sqrt{36}}\right)\)
\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}-...-\dfrac{1}{\sqrt{35}}+\dfrac{1}{\sqrt{36}}\right)\)
\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{1}+\dfrac{1}{6}\right)\)
\(\Leftrightarrow B< -\dfrac{5}{3}< 10=A\)
e,\(\sqrt{\dfrac{9}{169}}\)
f,\(\sqrt{1\dfrac{9}{16}}\)
g,\(\dfrac{\sqrt{2300}}{\sqrt{23}}\)
h,\(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
\(e,\sqrt{\dfrac{9}{169}}=\dfrac{\sqrt{9}}{\sqrt{169}}=\dfrac{\sqrt{3^2}}{\sqrt{13^2}}=\dfrac{3}{13}\)
\(f,\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{\sqrt{25}}{\sqrt{16}}=\dfrac{\sqrt{5^2}}{\sqrt{4^2}}=\dfrac{5}{4}\)
\(g,\dfrac{\sqrt{2300}}{\sqrt{23}}=\sqrt{\dfrac{2300}{23}}=\sqrt{100}=\sqrt{10^2}=10\)
\(h,\dfrac{\sqrt{12,5}}{\sqrt{0,5}}=\sqrt{\dfrac{12,5}{0,5}}=\sqrt{25}=\sqrt{5^2}=5\)
e, \(\sqrt{\dfrac{9}{169}}\)
\(=\sqrt{\dfrac{3^2}{13^2}}\)
\(=\dfrac{3}{13}\)
f, \(\sqrt{1\dfrac{9}{16}}\)
\(=\sqrt{\dfrac{25}{16}}\)
\(=\sqrt{\dfrac{5^2}{4^2}}\)
\(=\dfrac{5}{4}\)
g, \(\dfrac{\sqrt{2300}}{\sqrt{23}}\)
\(=\dfrac{10\sqrt{23}}{\sqrt{23}}\)
\(=10\)
h, \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
\(=\dfrac{\dfrac{5\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}}\)
\(=\dfrac{\dfrac{5\sqrt{2}}{2}\cdot2}{\sqrt{2}}\)
\(=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)
Cho P= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\) và Q= \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
a) rút gọn P
b) Tính M= P : Q và so sánh M với -1
\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\left(ĐKXĐ:x\ge0;x\ne9\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\sqrt{x}-3}{x-9}\)
\(b,M=P:Q\)
\(=\dfrac{-3\sqrt{x}-3}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
Ta thấy: \(\sqrt{x}\ge0\forall x\)
\(\Rightarrow\sqrt{x}+3\ge3\forall x\)
\(\Rightarrow\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\forall x\)
\(\Rightarrow\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{3}=-1\)
hay \(M\ge-1\)
#Toru