a) (-2)+ (-5) = -7

Vì: -7< -5

=> (-2)+ (-5) < -7

b) (-3)+ (-8)= -11

Vì: (-10) > (-11)

=> -10> (-3)+ (-8)

a)(-2)+(-5)<(-5)

b)(-10)>(-3)+(-8)

a) \(\left(-2\right)+\left(-5\right)..........\left(-5\right)\)

\(\left(-7\right)< \left(-5\right)\)

Vậy \(\left(-2\right)+\left(-5\right)< \left(-5\right)\)

b) \(\left(-10\right)...........\left(-3\right)+\left(-8\right)\)

\(\left(-10\right)>\left(-11\right)\)

Vậy \(\left(-10\right)>\left(-3\right)+\left(-8\right)\)

a, \(\left(-6\right)+\left(-3\right).....\left(-6\right)\)

\(\left(-9\right)< \left(-6\right)\)

Vậy \(\left(-6\right)+\left(-3\right)< \left(-6\right)\)

b,\(\left(-9\right)+\left(-12\right)....\left(-20\right)\)

\(\left(-21\right)< \left(-20\right)\)

\(\Rightarrow\left(-9\right)+\left(-12\right)< \left(-20\right)\)

a)

b)

a)

\(a\)) \(\left|3\right|< \left|5\right|\)

\(b\))\(\left|-3\right|< \left|-5\right|\)

\(c\)) \(\left|-1\right|>\left|0\right|\)

\(\left|2\right|=\left|-2\right|\)

\(A=\dfrac{\sqrt{\dfrac{9}{4}-3^{-1}+2018^0}}{25\%+1\dfrac{1}{4}-1,3}-\dfrac{\left(-\dfrac{1}{2}\right)^2-\sqrt{\dfrac{4}{9}}+0,4}{0,6-\dfrac{2}{3}.\left(-\dfrac{1}{4}-\dfrac{1}{2}\right)}\)

\(A=\dfrac{\sqrt{\dfrac{9}{4}-\dfrac{1}{3}+1}}{\dfrac{1}{4}+\dfrac{5}{4}-\dfrac{13}{10}}-\dfrac{\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{2}{5}}{\dfrac{3}{5}-\dfrac{2}{3}\left(-\dfrac{1}{4}-\dfrac{1}{2}\right)}\)

\(A=\dfrac{\sqrt{\dfrac{35}{12}}}{\dfrac{1}{5}}-\dfrac{-\dfrac{1}{60}}{\dfrac{11}{10}}\)

\(A=\dfrac{5\sqrt{105}}{6}+\dfrac{11}{66}\)

\(A=\dfrac{55\sqrt{105}}{66}+\dfrac{11}{66}\)

\(A=\dfrac{55\sqrt{105}+11}{66}\)

Điền đúng sai vào chỗ trống (....)

a) (−36):2=−18 Đ

b) 600:(−15)=−4 S

c) 27:(−1)=27 S

d) (−65):(−5)=13 Đ

\(a,x^2+20x+100=\left(x+10\right)^2\)

\(b,16x^2+24xy+9y^2=\left(4x+3y\right)^2\)

\(c,\left(2a+3b\right)\left(4a^2-6ab+9b^2\right)=8a^3+27a^3\)

\(d,\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)=125x^3-64y^3\)

a) (-5) . (-4) + (-5) . 14 = (-5 ) . [ (-4) + 14 ] = -50

b ) 3 . ( 5 + 8 ) = 13 . ( -3 ) + 13 . 8 = 65

\(\dfrac{5(x+y)}{2}=\dfrac{5(x+y)(x-y)}{2(x-y)} \\=\dfrac{5(x^2-y^2)}{2(x-y)}=\dfrac{5x^2-5y^2}{2x-2y}\)

Chỗ trống là: 2(x - y)

a) \(\left(x-1,3\right)^2=9\Leftrightarrow\left[{}\begin{matrix}x-1,3=3\\x-1,3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\)

b) 2^4-x = 32

⇔ 2^4-x = 2⁵

⇔ 4-x=5

⇔ x=-1

c) (x+1,5)²+(y-2,5)¹⁰=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,5\end{matrix}\right.\)

\(a,\left(x-1,3\right)^2=9\\ \Leftrightarrow\left(x-1,3+9\right)\left(x-1,3-9\right)=0\\ \Leftrightarrow\left(x-7,7\right)\left(x-10,3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7,7=\dfrac{77}{10}\\x=10,3=\dfrac{103}{10}\end{matrix}\right.\)

\(b,2^{4-x}=32=2^5\\ \Leftrightarrow4-x=5\\ \Leftrightarrow x=-1\)

\(c,\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1,5=-\dfrac{3}{2}\\y=2,5=\dfrac{5}{2}\end{matrix}\right.\)

a. (x - 1,3)² = 9

<=> (x - 1,3)² - 9 = 0

<=> (x - 1,3)² - 3² = 0

<=> (x - 1,3 - 3)(x - 1,3 + 3) = 0

<=> (x - 4,3)(x + 1,7) = 0

<=> \(\left[{}\begin{matrix}x-4,3=0\\x+1,7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\)