Em kiểm tra lại đề bài nhé  \(\frac{2}{x-y}\)hay \(\frac{2}{x-2}\)

dạ là x-y ạ

Bài 5 hình 1: (tự vẽ hình nhé bạn)
a) Xét ΔABD và ΔACB ta có:
\(\widehat{BAD}\)= \(\widehat{BAC}\) (góc chung)
\(\widehat{ABD}\)= \(\widehat{ACB}\) (gt)
=> ΔABD ~ ΔACB (g-g)
=> \(\dfrac{AB}{AC}\) = \(\dfrac{BD}{CB}\) = \(\dfrac{AD}{AB}\) (tsđd)
b) Ta có: \(\dfrac{AB}{AC}\) = \(\dfrac{AD}{AB}\) (cm a)
=> \(AB^2\) = AD.AC
=> \(2^2\) = AD.4
=> AD = 1 (cm)
Ta có: AC = AD + DC (D thuộc AC)
      => 4   =   1   + DC
      => DC = 3 (cm)
c) Xét ΔABH và ΔADE ta có: 
   \(\widehat{AHB}\) = \(\widehat{AED}\) (=\(90^0\))
   \(\widehat{ADB}\) = \(\widehat{ABH}\) (ΔABD ~ ΔACB)
=> ΔABH ~ ΔADE
=> \(\dfrac{AB}{AD}\) = \(\dfrac{AH}{AE}\) = \(\dfrac{BH}{DE}\) (tsdd)
Ta có: \(\dfrac{S_{ABH}}{S_{ADE}}\) = \(\left(\dfrac{AB}{AD}\right)^2\)= \(\left(\dfrac{2}{1}\right)^2\)= 4
=> đpcm

Tiếp bài 5 hình 2 (tự vẽ hình)
a) Xét ΔABC vuông tại A ta có:
\(BC^2\) = \(AB^2\) + \(AC^2\)
\(BC^2\) = \(21^2\) + \(28^2\)
BC = 35 (cm)
b) Xét ΔABC và ΔHBA ta có:
\(\widehat{BAC}\) = \(\widehat{AHB}\) ( =\(90^0\))
\(\widehat{ABC}\) = \(\widehat{ABH}\) (góc chung)
=> ΔABC ~ ΔHBA (g-g)
=> \(\dfrac{AB}{BH}\) = \(\dfrac{BC}{AB}\) (tsdd)
=> \(AB^2\) = BH.BC
=> \(21^2\) = 35.BH
=> BH = 12,6 (cm)
c) Xét ΔABC ta có:
BD là đường p/g (gt)
=> \(\dfrac{AD}{DC}\) = \(\dfrac{AB}{BC}\) (t/c đường p/g)
Xét ΔABH ta có: 
BE là đường p/g (gt)
=> \(\dfrac{HE}{AE}\) = \(\dfrac{BH}{AB}\) (t/c đường p/g)
Mà: \(\dfrac{AB}{BC}\) = \(\dfrac{BH}{AB}\) (cm b)
=> đpcm
d) Ta có: \(\left\{{}\begin{matrix}\widehat{HBE}+\widehat{BEH}=90^0\\\widehat{ABD}+\widehat{ADB=90^0}\\\widehat{HBE}=\widehat{ABD}\end{matrix}\right.\)
=> \(\widehat{BEH}=\widehat{ADB}\)
Mà \(\widehat{BEH}=\widehat{AED}\) (2 góc dd)
Nên \(\widehat{ADB}=\widehat{AED}\)
=> đpcm

(oh) hóa trị 1 mà zn hóa trị 2=> cthh la zn(oh)2

với lại ko có oh2 dau chi co OH hoac la H2O

phải viết là Zn(OH)₂ vì nhóm (OH) hóa trị I

Key : 133 ; 322 ; 329 ; 266 ; 455 ; 644 ; 833 ; 714......

Đây chỉ vài vd 

#Sun

Bạn có thể giải chi tiết ko?

ĐKXĐ: \(\begin{cases}x<>0\\ x<>-1\\ -x^2+3x+1<>0\end{cases}\)

=>\(\begin{cases}x\notin\left\lbrace0;-1\right\rbrace\\ x^2-3x-1<>0\end{cases}\Rightarrow\begin{cases}x\notin\left\lbrace0;-1\right\rbrace\\ x^2-3x+\frac94<>\frac{13}{4}\end{cases}\)

=>\(\begin{cases}x\notin\left\lbrace0;-1\right\rbrace\\ \left(x-\frac32\right)^2<>\frac{13}{4}\end{cases}\Rightarrow\begin{cases}x\notin\left\lbrace0;-1\right\rbrace\\ x-\frac32<>\pm\frac{\sqrt{13}}{2}\end{cases}\)

=>\(x\notin\left\lbrace0;-1;\frac{3-\sqrt{13}}{2};\frac{3+\sqrt{13}}{2}\right\rbrace\)

a: \(D=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x-x^2+1}{3x}\)

\(=\frac{\left(x+2\right)\left(x+1\right)+6x-3\cdot3x\cdot\left(x+1\right)}{3x\left(x+1\right)}\cdot\frac{x+1}{2-4x}+\frac{x^2-3x-1}{3x}\)

\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(2-4x\right)}+\frac{x^2-3x-1}{3x}\)

\(=\frac{-8x^2+2}{3x\left(-4x+2\right)}+\frac{x^2-3x-1}{3x}=\frac{2\left(4x^2-1\right)}{3x\cdot2\cdot\left(2x-1\right)}+\frac{x^2-3x-1}{3x}\)

\(=\frac{2x+1}{3x}+\frac{x^2-3x-1}{3x}=\frac{x^2-3x-1+2x+1}{3x}=\frac{x^2-x}{3x}=\frac{x-1}{3}\)

b: Khi x=2010 thì \(D=\frac{2010-1}{3}=\frac{2009}{3}\)

c: D<0

=>x-1<0

=>x<1

=>x<1 và x∉{0;-1;\(\frac{3-\sqrt{13}}{2}\) }

\(1,\\ a,\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow\left(x-7\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ d,\Leftrightarrow\left(2x+3\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\\ 2,\\ a,\Leftrightarrow\left(x+5\right)^2=0\Leftrightarrow x=-5\\ b,\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\\ c,\Leftrightarrow\left(x-9\right)^2=0\Leftrightarrow x=9\\ d,\Leftrightarrow\left(x-3\right)^3=0\Leftrightarrow x=3\\ e,\Leftrightarrow3x\left(x^2-2x+3\right)=0\\ \Leftrightarrow3x\left(x^2-2x+1+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2+2=0\left(vô.nghiệm\right)\end{matrix}\right.\\ \Leftrightarrow x=0\)

\(f,\Leftrightarrow3x\left(x^2-4x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Bài 1:

a) \(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

b) \(\Rightarrow3x\left(x-4\right)-\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-7\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\)

d) \(\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Bài 2:

a) \(\Rightarrow\left(x+5\right)^2=0\Rightarrow x=-5\)

b) \(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x=\dfrac{1}{2}\)

c) \(\Rightarrow\left(x-9\right)^2=0\Rightarrow x=9\)

d) \(\Rightarrow\left(x-3\right)^3=0\Rightarrow x=3\)

e) \(\Rightarrow3x\left(x^2-6x+9\right)=0\)

\(\Rightarrow3x\left(x-3\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

f) \(\Rightarrow3x\left(x^2-4x+4\right)=0\)

\(\Rightarrow3x\left(x-2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Mỗi giờ xe thứ hai đi nhanh hơn xe thứ nhất là : \(35-30=5\) ( km )

Có 7h30p - 6h = 1h30p là thời gian xe một xuất phát trước xe hai. Đổi 1h30p thành 1,5 giờ

Trong 1h30p xe thứ nhất đi được là : \(30\times1,5=45\left(km\right)\)

Từ trên ta có mỗi giờ khoảng cách giữa 2 xe giảm đi 5km

=> Số giờ sau 2 xe có thể gặp nhau : \(45\div5=9\left(h\right)\)

Đáp số :................

1: \(35^2=\left(30+5\right)^2=30^2+2\cdot30\cdot5+5^2\)

=900+300+25

=1225

2: \(25^2=\left(20+5\right)^2=20^2+2\cdot20\cdot5+5^2\)

=400+200+25

=625

3: \(75^2=\left(100-25\right)^2\)

\(=100^2-2\cdot100\cdot25+25^2\)

=10000-5000+625

=5625

4: \(95^2=\left(100-5\right)^2\)

\(=100^2-2\cdot100\cdot5+5^2\)

=10000-1000+25

=9025

5: \(101\cdot99=\left(100-1\right)\left(100+1\right)=100^2-1\)

=10000-1

=9999

6: \(36\cdot44=\left(40-4\right)\left(40+4\right)\)

\(=40^2-4^2\)

=1600-16

=1584

7: \(72\cdot68=\left(70+2\right)\left(70-2\right)\)

\(=70^2-2^2\)

=7000-4

=6996

8: \(\left(625^2+3\right)\left(25^4-3\right)-5^{16}+10\)

\(=\left(25^4+3\right)\left(25^4-3\right)-25^8+10\)

\(=25^8-9-25^8+10\)

=10-9

=1

9: \(39^2+78\cdot61+61^2\)

\(=39^2+2\cdot39\cdot61+61^2\)

\(=\left(39+61\right)^2=100^2=10000\)

10: \(50^2-49\cdot51\)

\(=50^2-\left(50-1\right)\left(50+1\right)\)

\(=50^2-\left(50^2-1\right)=1\)

11: \(\frac{41^2+39^2+82\cdot39}{41^1-39^2}=\frac{\left(41+39\right)^2}{\left(41-39\right)\left(41+39\right)}=\frac{41+39}{41-39}=\frac{80}{2}=40\)

12: \(127^2+146\cdot127+73^2\)

\(=127^2+2\cdot127\cdot73+73^2\)

\(=\left(127+73\right)^2=200^2=40000\)