1+3+5+...+x=36
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: Hai cạnh đáy là AB,CD
Hai cạnh bên là AD,BC
b: Các cặp góc kề cạnh đáy là:
\(\widehat{BAD};\widehat{ABC}\)
\(\widehat{ADC};\widehat{BCD}\)
Các cặp góc kề cạnh bên là:
\(\widehat{BAD};\widehat{ADC}\)
\(\widehat{ABC};\widehat{BCD}\)
c: Hai đường chéo là AC,BD
Bài 2:
a: Ta có: ΔDAC vuông cân tại D
=>\(\widehat{DAC}=\widehat{DCA}=45^0\)
Ta có: ΔABC vuông cân tại A
=>\(\widehat{ABC}=\widehat{ACB}=45^0\)
Ta có: \(\widehat{DAC}=\widehat{ACB}\left(=45^0\right)\)
mà hai góc này là hai góc ở vị trí so le trong
nên AD//CB
=>ABCD là hình thang
Hình thang ABCD có AD\(\perp\)DC
nên ABCD là hình thang vuông
b: ABCD là hình thang vuông có hai đáy là AD,CB và AD\(\perp\)DC
=>CB\(\perp\)CD
=>\(\widehat{ADC}=\widehat{DCB}=90^0\)
Ta có: AD//CB
=>\(\widehat{DAB}+\widehat{ABC}=180^0\)
=>\(\widehat{DAB}=180^0-45^0=135^0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\\ =\dfrac{8}{9}-\left(\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\\ =\dfrac{8}{9}-\left(\dfrac{1}{8\cdot9}+\dfrac{1}{7\cdot8}+\dfrac{1}{6\cdot7}+\dfrac{1}{6\cdot5}+\dfrac{1}{4\cdot5}+\dfrac{1}{3\cdot4}+\dfrac{1}{2\cdot3}+\dfrac{1}{1\cdot2}\right)\\ =\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\\ =\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\\ =\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)
b)
\(\left(-\dfrac{1}{2}\right)-\left(\dfrac{-3}{5}\right)+\left(-\dfrac{1}{9}\right)+\dfrac{1}{127}-\dfrac{7}{18}+\dfrac{4}{35}-\left(\dfrac{-2}{7}\right)\\ =\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{127}\\ =\dfrac{-9-2-7}{18}+\dfrac{21+10+4}{35}+\dfrac{1}{127}\\ =-1+1+\dfrac{1}{127}\\ =\dfrac{1}{127}\)
c) (*sửa*)
\(\dfrac{3}{5}+\dfrac{3}{11}-\dfrac{-3}{7}+\dfrac{2}{97}-\dfrac{1}{35}-\dfrac{3}{4}-\dfrac{23}{44}\\ =\dfrac{3}{5}+\dfrac{3}{11}+\dfrac{3}{7}+\dfrac{2}{97}-\dfrac{1}{35}-\dfrac{3}{4}+\dfrac{23}{44}\\ =\left(\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{1}{35}\right)+\left(\dfrac{3}{11}-\dfrac{3}{4}-\dfrac{23}{44}\right)+\dfrac{2}{97}\\ =\dfrac{21+15-1}{35}+\dfrac{12-33-23}{44}+\dfrac{2}{97}\\ =1+\left(-1\right)+\dfrac{2}{97}\\ =\dfrac{2}{97}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2^2+2^4+2^6+...+2^{200}\\ 2^2A=2^4+2^6+...+2^{202}\\ 4A-A=\left(2^4+2^6+2^8+...+2^{202}\right)-\left(2^2+2^4+2^6+...+2^{200}\right)\\ 3A=2^{202}-2^2\)
\(=>3A+4=2^{202}-2^2+4=2^{202}-4+4=2^{202}\)
\(=>2^{202}=4^n\\ =>2^{202}=\left(2^2\right)^n\\ =>2^{202}=2^{2n}\\ =>2n=202\\ =>n=101\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ABCD là hình vuông
=>AB//CD
mà C\(\in\)DE
nên AB//DE
Ta có: DEFG là hình chữ nhật
=>DE//FG
mà AB//DE
nên AB//FG
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(x-2\right)\left(3x-1\right)\left(x^2-4x+1\right)\)
\(=\left(3x^2-x-6x+2\right)\left(x^2-4x+1\right)\)
\(=\left(3x^2-7x+2\right)\left(x^2-4x+1\right)\)
\(=3x^4-12x^3+3x^2-7x^3+28x^2-7x+2x^2-8x+2\)
\(=3x^4-19x^3+33x^2-15x+2\)
b: \(x\left(3-4x\right)\left(2x^2-3x\right)\)
\(=\left(-4x^2+3x\right)\left(2x^2-3x\right)\)
\(=-8x^4+12x^3+6x^3-9x^2\)
\(=-8x^4+18x^3-9x^2\)
a)
\(\left(x-2\right)\left(3x-1\right)\left(x^2-4x+1\right)\\ =\left(3x^2-6x-x+2\right)\left(x^2-4x+1\right)\\ =\left(3x^2-7x+2\right)\left(x^2-4x+1\right)\\ =3x^4-12x^3+3x^2-7x^3+28x^2-7x-8x+2\\ =3x^4-19x^3+31x^2-15x+2\)
b)
\(x\left(3-4x\right)\left(2x^2-3x\right)\\ =\left(3x-4x^2\right)\left(2x^2-3x\right)\\ =6x^3-9x^2-8x^4+12x^3\\ =-8x^4+18x^3-9x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(32< 2^x< 128\\ =>2^5< 2^x< 2^7\\ =>5< x< 7\\ =>x=6\)
b)
\(2\cdot16\ge2^x>4\\ =>2\cdot2^4\ge2^x>2^2\\ =>2^5\ge2^x>2^2\\ =>5\ge x>2\\ =>x\in\left\{3;4;5\right\}\)
c)
\(9\cdot27\le3^x\le243\\ =>3^2\cdot3^3\le3^x\le3^5\\ =>3^5\le3^x\le3^5\\ =>5\le x\le5\\ =>x=5\)
d)
\(x^{2019}=x\\ =>x^{2019}-x=0\\ =>x\left(x^{2018}-1\right)=0\)
TH1: x = 0
TH2: `x^2018-1=0`
`=>x^2018=1`
`=>x^2018=1^2018`
`=>x=1` hoặc `x=-1`
a: \(32< 2^x< 128\)
=>\(2^5< 2^x< 2^7\)
=>5<x<7
mà x là số tự nhiên
nên x=6
b: \(2\cdot16>=2^x>4\)
=>\(2^5>=2^x>2^2\)
=>2<x<=5
mà x là số tự nhiên
nên \(x\in\left\{3;4;5\right\}\)
c: \(9\cdot27< =3^x< =243\)
=>\(243< =3^x< =243\)
=>\(3^x=243=3^5\)
=>x=5
d: \(x^{2019}=x\)
=>\(x\left(x^{2018}-1\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x^{2018}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^{2018}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
e: \(2^{x+1}+4\cdot2^x=3\cdot2^7\)
=>\(2^x\cdot2+4\cdot2^x=6\cdot2^6\)
=>\(6\cdot2^x=6\cdot2^6\)
=>x=6
f: \(2^{2x}+2^{2x+3}=3^2\cdot8^4\)
=>\(2^{2x}+2^{2x}\cdot8=9\cdot8^4\)
=>\(9\cdot2^{2x}=9\cdot2^{12}\)
=>2x=12
=>x=6
g: \(27^{x+1}=9^{x+5}\)
=>\(3^{3\left(x+1\right)}=3^{2\left(x+5\right)}\)
=>3(x+1)=2(x+5)
=>3x+3=2x+10
=>3x-2x=10-3
=>x=7
h: \(3^{x+2}+5\cdot3^{x+1}=648\)
=>\(3^x\cdot9+5\cdot3^x\cdot3=648\)
=>\(3^x\cdot24=648\)
=>\(3^x=\dfrac{648}{24}=27=3^3\)
=>x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Thay x=2 vào ta có:
\(2^2-4m\cdot2+1=0\\ \Leftrightarrow4-8m+1=0\\ \Leftrightarrow5-8m=0\\ \Leftrightarrow8m=5\\ \Leftrightarrow m=\dfrac{5}{8}\)
b) Thay x=2 vào ta có:
\(3\cdot2^2-5m\cdot2+7\\ \Leftrightarrow12-10m+7=0\\ \Leftrightarrow19-10m=0\\ \Leftrightarrow10m=19\\\Leftrightarrow m=\dfrac{19}{10}\)
a:
Đặt \(x^2-4mx+1=0\left(1\right)\)
Thay x=2 vào (1), ta được:
\(2^2-4m\cdot2+1=0\)
=>\(4-8m+1=0\)
=>5-8m=0
=>8m=5
=>\(m=\dfrac{5}{8}\)
b: Đặt \(3x^2-5mx+7=0\left(2\right)\)
Thay x=2 vào (2), ta được:
\(3\cdot2^2-5m\cdot2+7=0\)
=>12-10m+7=0
=>19-10m=0
=>10m=19
=>\(m=\dfrac{19}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tỉ số giữa Số học sinh khá và cả lớp là:
\(\dfrac{1}{2}\left(1-\dfrac{1}{4}\right)=\dfrac{1}{2}\cdot\dfrac{3}{4}=\dfrac{3}{8}\)
Số học sinh trung bình chiếm: \(1-\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{5}{8}-\dfrac{2}{8}=\dfrac{3}{8}\)(số học sinh cả lớp)
Số học sinh cả lớp là \(24:\dfrac{3}{8}=24\cdot\dfrac{8}{3}=64\left(bạn\right)\)
\(1+3+5+...+x=36\\\left[ \left(x-1\right):2+1\right]\cdot\left(x+1\right):2=36\\ \dfrac{x-1+2}{2}\cdot\dfrac{x+1}{2}=36\\ \dfrac{x+1}{2}\cdot\dfrac{x+1}{2}=36\\ \dfrac{\left(x+1\right)^2}{4}=36\\ \left(x+1\right)^2=36\cdot4=144\)
TH1: x + 1 = 12 => x = 11
TH2: x + 1 = -12 => x = -13
Vì: x phải lớn hơn 0 => x = 11
1+3+5+...+x =36
1+3+5+7+9+11 =36