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c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\)
Ở nơi x=9/4-1/2 là x-9/4-1/2 nha
a. -1,5 + 2x = 2,5
<=> 2x = 2,5 + 1,5
<=> 2x = 4
<=> x = 2
b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)
<=> 9x + 45 - 3 = 8
<=> 9x = 8 + 3 - 45
<=> 9x = -34
<=> x = \(\dfrac{-34}{9}\)
3:
a: C=3x^2+5y^3+2
D=3x^2+4y^3-3/4
Bậc của C là 3
b: Khi x=-1 và y=1 thì D=3+4-3/4=7-3/4=25/4
c: C-D
=3x^2+5y^3+2-3x^2-4y^3+3/4
=y^3+11/4
Do \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Rightarrow1-\dfrac{b}{a}=1-\dfrac{d}{c}\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\) (đpcm)
cặp : Ea// Fb (vì góc e +góc f =180 mà 2 góc này ở vị trí trong cùng phía)
cặp Fb // DC (vì có góc F = góc D (=110) mà 2 góc này ở vị trí đồng vị)
cặp : Ea //DC vì Ea // Fb, Fb //DC (tính chất bắc cầu)
\(\\ \)
Câu 4:
a: Xét ΔABD và ΔAED có
AB=AE
\(\widehat{BAD}=\widehat{EAD}\)
AD chung
Do đó: ΔABD=ΔAED
Câu 1:
\(a,=\dfrac{1}{2}+9\cdot\dfrac{1}{9}-18=\dfrac{1}{2}+1-18=-\dfrac{33}{2}\\ b,=2-1+4\cdot\dfrac{1}{4}+9\cdot\dfrac{1}{9}\cdot9=1+1+9=11\\ c,=-21,3\left(54,6+45,4\right)=-21,3\cdot100=-2130\\ d,B=\left(\dfrac{1}{16}+\dfrac{1}{2}-\dfrac{1}{16}\right):\left(\dfrac{1}{8}-\dfrac{1}{8}+1\right)=\dfrac{1}{2}:1=\dfrac{1}{2}\)
bài 2
a)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(3+5x\right)\left(x-y\right)\)
c)
\(=x\left(x-1\right)+y\left(x-1\right)=\left(x+y\right)\left(x-1\right)\)
d)
\(=7x\left(5x-y\right)+2\left(5x-y\right)+3y\left(5x-y\right)\)
\(=\left(7x+2+3y\right)\left(5x-y\right)\)
e)
=\(2y\left(3-x\right)-3xy\left(3-x\right)=\left(2-3x\right)\left[y\left(3-x\right)\right]\)
bài 1
a) = x(5x-6)
b) = 3x(y+4x)
c)=6x2(4x+1)
d)=\(2ab\left(a-2\right)\)
e)\(=\left(x-5y\right)\left(x+y\right)\)
f) \(\left(x-y\right)\left(y+2\right)\)
1) \(a^3+1=\left(a+1\right)\left(a^2-a+1\right)\)
2) \(a^3+6a^2+12a+8=\left(a+2\right)^3\)
3) \(8a^3+12a^2+6a+1=\left(2a+1\right)^3\)
4) \(27a^3-1=\left(3a-1\right)\left(9a^2+3a+1\right)\)
5) \(x^3-12x^2+48x-64=\left(x-4\right)^3\)
6) \(27a^3-54a^2+36a-8=\left(3a-2\right)^3\)