Thực hiện phép tính: \(\left( { - 2} \right).29 + \left( { - 2} \right).\left( { - 99} \right)\)\( + \left( { - 2} \right).\left( { - 30} \right)\)
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a) \(4+7=11\)
b) \(\left( { - 4} \right)\) và \(\left( { - 7} \right)\) là hai số nguyên âm có số đối lần lượt là 4 và 7 nên \(\left( { - 4} \right) + \left( { - 7} \right) = - \left( {4 + 7} \right) = - 11\).
c) \(\left( { - 99} \right)\) có số đối là 99
\(\left( { - 11} \right)\) có số đối là 11.
Vậy \(\left( { - 99} \right) + \left( { - 11} \right) = - \left( {99 + 11} \right) = - 110\)
d) \(\left( { + 99} \right) + \left( { + 11} \right) = 99 + 11 = 110\)
e) \(\left( { - 65} \right) + \left( { - 35} \right) = - \left( {65 + 35} \right) = - 100\)
Mk lm đc câu a thôi nhé !
A= 150-(100-99+98-97+...-3+2-1)
từ 1-100 có 100 SH. Ta nhóm 4 số vs nhau như sau : (100-00+98-87)+(...)+(4-3+2-1)
Có tất cả số nhóm là : 100:4=25 nhóm. Mà mỗi nhóm ta tính có kết quả là 2, vậy tao có
A=150-(2.25)
A=150-50
A=100
\(a,\cdot\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}\right)^2\right]\right\}:\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}\\ =\left[\left(8:2,4\right)\cdot\left(5,25:7\right)\right]:\left[\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)\right]\\ =\left(\dfrac{10}{3}\cdot\dfrac{3}{4}\right):\left(3:\dfrac{9}{2}\right)\\ =\dfrac{5}{2}:\dfrac{2}{3}\\ =\dfrac{15}{4}\)
a: \(\dfrac{\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}^2\right)\right]\right\}}{\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}}\)
\(=\dfrac{\dfrac{8}{2,4}\cdot\dfrac{5,25}{7}}{\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{3}{4}}{3:\left(4\cdot\dfrac{9}{8}\right)}\)
\(=\dfrac{\dfrac{10}{4}}{3:\left(\dfrac{9}{2}\right)}=\dfrac{5}{2}:\left(3\cdot\dfrac{2}{9}\right)=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{15}{4}\)
b: \(\sqrt{\left(x-\sqrt{2}\right)^2}=\left|x-\sqrt{2}\right|>=0\forall x\)
\(\sqrt{\left(y+\sqrt{2}\right)^2}=\left|y+\sqrt{2}\right|>=0\forall y\)
\(\left|x+y+z\right|>=0\forall x,y,z\)
Do đó: \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|>=0\forall x,y,z\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{matrix}\right.\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+.....+\frac{1}{\left(x+99\right)\left(x+100\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+.....+\frac{1}{x+99}-\frac{1}{x+100}\)
\(=\frac{1}{x}-\frac{1}{x+100}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{1}{x}-\frac{1}{x+100}=\frac{x+100-x}{x\left(x+100\right)}=\frac{100}{x\left(x+100\right)}\)
(8 - 5x).(x + 2) + 4.(x - 2)(x - 1) + 2.(x - 2)(x + 2) + 10
= (8x + 16 - 5x2 - 10x) + 4.(x2 - 3x + 2) + 2.(x2 - 4) + 10
= 8x + 16 - 5x2 - 10x + 4x2 - 12x + 8 + 2x2 - 8 + 10
= (8x - 10x - 12x) + (-5x2 + 4x2 + 2x2) + (16 + 8 - 8 + 10)
= -14x + x2 + 26
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)
=1/x-1/x+2014
\(=\dfrac{x+2014-x}{x\left(x+2014\right)}=\dfrac{2014}{x\left(x+2014\right)}\)
$=(6x^2+x-2)(3-x)$
$=18x^2-6x^3+3x-x^2-6+2x$
$=-6x^3+17x^2+5x-6$
\(=\dfrac{5\left(x-2y\right)^4+\left(x-2y\right)^2-\left(x-2y\right)}{x-2y}\)
=5(x-2y)^3+(x-2y)-1
\(\left( { - 2} \right).29 + \left( { - 2} \right).\left( { - 99} \right)\)\( + \left( { - 2} \right).\left( { - 30} \right)\)\( = \left( { - 2} \right)\left( {29 - 99 - 30} \right)\)\( = \left( { - 2} \right).\left( { - 100} \right) = 200\)