Điền dấu " <, >" vào chỗ trống cho đúng :
a) \(\left(0,6\right)^2........\left(0,6\right)\)
b) \(\left(1,3\right)^2.........1,3\)
Điền dấu ">", "<" thích hợp vào chỗ trống :
a) \(\left(-2\right)+\left(-5\right)......\left(-5\right)\)
b) \(\left(-10\right)......\left(-3\right)+\left(-8\right)\)
a) (-2)+ (-5) = -7
Vì: -7< -5
=> (-2)+ (-5) < -7
b) (-3)+ (-8)= -11
Vì: (-10) > (-11)
=> -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 \(\left(-10\right)>\left(-3\right)+\left(-8\right)\)
Điền dấu > , < vào chỗ trống :
a) \(\left(-6\right)+\left(-3\right).....\left(-6\right)\)
b) \(\left(-9\right)+\left(-12\right).....\left(-20\right)\)
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,\(\left(-9\right)+\left(-12\right)....\left(-20\right)\)
\(\left(-21\right)< \left(-20\right)\)
\(\Rightarrow\left(-9\right)+\left(-12\right)< \left(-20\right)\)
Điền dấu \(>,=,< \) vào chỗ trống :
a) \(\left|3\right|......\left|5\right|\)
b) \(\left|-3\right|.....\left|-5\right|\)
c) \(\left|-1\right|......\left|0\right|\)
d) \(\left|2\right|.....\left|-2\right|\)
\(a\)) \(\left|3\right|< \left|5\right|\)
\(b\))\(\left|-3\right|< \left|-5\right|\)
\(c\)) \(\left|-1\right|>\left|0\right|\)
\(\left|2\right|=\left|-2\right|\)
\(\dfrac{\sqrt{\dfrac{9}{4}-3^{-1}+2018^0}}{25\%+1\dfrac{1}{4}-1,3}-\dfrac{\left(\dfrac{-1}{2}\right)^2-\sqrt{\dfrac{4}{9}}+0,4}{0,6-\dfrac{2}{3}.\left(\dfrac{-1}{4}-\dfrac{1}{2}\right)}\)
\(A=\dfrac{\sqrt{\dfrac{9}{4}-3^{-1}+2018^0}}{25\%+1\dfrac{1}{4}-1,3}-\dfrac{\left(-\dfrac{1}{2}\right)^2-\sqrt{\dfrac{4}{9}}+0,4}{0,6-\dfrac{2}{3}.\left(-\dfrac{1}{4}-\dfrac{1}{2}\right)}\)
\(A=\dfrac{\sqrt{\dfrac{9}{4}-\dfrac{1}{3}+1}}{\dfrac{1}{4}+\dfrac{5}{4}-\dfrac{13}{10}}-\dfrac{\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{2}{5}}{\dfrac{3}{5}-\dfrac{2}{3}\left(-\dfrac{1}{4}-\dfrac{1}{2}\right)}\)
\(A=\dfrac{\sqrt{\dfrac{35}{12}}}{\dfrac{1}{5}}-\dfrac{-\dfrac{1}{60}}{\dfrac{11}{10}}\)
\(A=\dfrac{5\sqrt{105}}{6}+\dfrac{11}{66}\)
\(A=\dfrac{55\sqrt{105}}{66}+\dfrac{11}{66}\)
\(A=\dfrac{55\sqrt{105}+11}{66}\)
Điền đúng sai vào chỗ trống (....)
a) \(\left(-36\right):2=-18.....\)
b) \(600:\left(-15\right)=-4.....\)
c) \(27:\left(-1\right)=27......\)
d) \(\left(-65\right):\left(-5\right)=13....\)
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
Điền vào chỗ trống để được đẳng thức đúng :
a) \(x^2+20x+...=\left(...+...\right)^2\)
b) \(16x^2+24xy+...=\left(...+...\right)^2\)
c) \(\left(2a+3b\right)\left(...-...+...\right)=8a^3+27a^3\)
d) \(\left(5x-...\right)\left(...+20xy+...\right)=125x^3-64y^3\)
Pleasehelpme =))
Ainhanhtickcho =))
Thanks =))
\(a,x^2+20x+100=\left(x+10\right)^2\)
\(b,16x^2+24xy+9y^2=\left(4x+3y\right)^2\)
\(c,\left(2a+3b\right)\left(4a^2-6ab+9b^2\right)=8a^3+27a^3\)
\(d,\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)=125x^3-64y^3\)
Điền số thích hợp vào chỗ trống (....)
a) \(\left(-5\right).\left(-4\right)+\left(-5\right).14=\left(-5\right).\left[\left(-4\right)+....\right]=.....\)
b) \(3.\left(....+8\right)=13.\left(-3\right)+13.......=65\)
a) (-5) . (-4) + (-5) . 14 = (-5 ) . [ (-4) + 14 ] = -50
b ) 3 . ( 5 + 8 ) = 13 . ( -3 ) + 13 . 8 = 65
Điền vào chỗ trống để có kết quả đúng
\(\dfrac{5\left(x+y\right)}{2}=\dfrac{5x^2-5y^2}{........}\)
\(\dfrac{5(x+y)}{2}=\dfrac{5(x+y)(x-y)}{2(x-y)} \\=\dfrac{5(x^2-y^2)}{2(x-y)}=\dfrac{5x^2-5y^2}{2x-2y}\)
Bài 1:
a) \(\left(x-1,3\right)^2=9\)
b) \(2^{4-x}=32\)
c) \(\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\)
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) 24-x = 32
⇔ 24-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
<=> \(\left[{}\begin{matrix}x-4,3=0\\x+1,7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\)