a) \(56+\left(-29\right)+\left(-7\right)+28+13+\left(-35\right)\)
b) \(\left(-213\right)+186+\left(-14\right)+217+54+\left(-49\right)\)
c)\(435+\left(-43\right)+\left(-483\right)+\left(-57\right)+383+\left(-415\right)\)
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435 + ( - 43 ) + ( - 487 ) + ( - 57 ) + 383 + ( - 415 )
= 435 + ( - 415 ) + ( - 43 ) + 383 + ( - 487 ) + ( - 57 )
= [ 435 + ( - 415 ) ] + [ ( - 43 ) + 383 ] + [ ( - 487 ) + ( - 57 ) ]
= 20 + 340 + ( - 544 )
= 360 + ( - 544 )
= - 184
1316 + 317 + ( - 1213 ) + ( - 314 + ( - 85 ) )
= [ 1316 + ( - 1213 ) ] + [ 317 + ( - 314 + ( - 85 ) ) ]
= 103 + ( - 82 )
= 21
1 + ( - 4 ) + 7 + ( - 10 ) + 13 + ( - 16 )
= [ 1 + ( - 10 ) ] + [ ( - 4 ) + ( - 16 ) ] + [ 7 + 13 ]
= ( - 9 ) + ( - 20 ) + 20
= ( - 9 ) + [ ( - 20 ) + 20 ]
= ( - 9 ) + 0
= - 9
a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)
b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)
c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)
\(=\frac{1}{x+3}-\frac{1}{x+34}\)
\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)
\(\Rightarrow x=31\)
Vậy, x = 31
Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với \(x,k\inℝ;x\ne0;x\ne-k\)
Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
a) 23 + 45 = 68
b) \(\left( { - 42} \right) + \left( { - 54} \right) = - \left( {42 + 54} \right) = - 96\)
c) \(2025 + \left( { - 2025} \right) = 0\) vì 2025 và \( - 2025\) là 2 số đối nhau.
d) \(15 + \left( { - 14} \right) = 15 - 14 = 1\);
e) \(35 + \left( { - 135} \right) = - \left( {135 - 35} \right) = - 100\)
b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)
\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)
\(=4\left(7+3\sqrt{5}\right)\)
\(=28+12\sqrt{5}\)
Lời giải:
a.
$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$
$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$
$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$
$=2\sqrt{5}-5\sqrt{10}$
$\Rightarrow A=\sqrt{10}-5\sqrt{5}$
b.
$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$
$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$
$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$
$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$
$\Rightarrow B=28+12\sqrt{5}$
c.
$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$
$=(7-5)(6-\sqrt{35})$
$=2(6-\sqrt{35})=12-2\sqrt{35}$
a) \(56+\left(-29\right)+\left(-7\right)+28+13+\left(-35\right)\)
\(=27+\left(-7\right)+28+13+\left(-35\right)\)
\(=20+28+13+\left(-35\right)\)
\(=48+13+\left(-35\right)\)
\(=61+\left(-35\right)\)
\(=26\)
b) \(\left(-213\right)+186+\left(-14\right)+217+54+\left(-49\right)\)
\(=\left(-27\right)+\left(-14\right)+217+54+\left(-49\right)\)
\(=\left(-41\right)+217+54+\left(-49\right)\)
\(=176+54+\left(-49\right)\)
\(=230+\left(-49\right)\)
\(=181\)
c) \(435+\left(-43\right)+\left(-483\right)+\left(-57\right)+383+\left(-415\right)\)
\(=392+\left(-483\right)+\left(-57\right)+383+\left(-415\right)\)
\(=\left(-91\right)+\left(-57\right)+383+\left(-415\right)\)
\(=\left(-148\right)+383+\left(-415\right)\)
\(=235+\left(-415\right)\)
\(=-180\)