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\(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+....+\dfrac{7}{86\cdot93}=\dfrac{\overline{a1}}{\overline{bcd}}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{86}-\dfrac{1}{93}=\dfrac{\overline{a1}}{\overline{bcd}}\)
\(\Rightarrow\dfrac{1}{2}-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-\left(\dfrac{1}{16}-\dfrac{1}{16}\right)-...-\left(\dfrac{1}{86}-\dfrac{1}{86}\right)-\dfrac{1}{93}=\dfrac{\overline{a1}}{\overline{bcd}}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{93}=\dfrac{\overline{a1}}{\overline{bcd}}\)
\(\Rightarrow\dfrac{91}{186}=\dfrac{\overline{a1}}{\overline{bcd}}\)
(1): \(\overline{a1}=91\Rightarrow a=9\)
(2): \(\overline{bcd}=186\Rightarrow\left\{{}\begin{matrix}b=1\\c=8\\d=6\end{matrix}\right.\)
Vậy: ...
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Độ dài chiều cao mặt bên của hình chóp tứ giác đều:
60 : 4 : 6 . 2 = 5 (cm)
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\(9\left(x-3y\right)^2-25\left(2x+y\right)^2\)
\(=\left[3\left(x-3y\right)\right]^2-\left[5\left(2x+y\right)\right]^2\)
\(=\left(3x-9y\right)^2-\left(10x+5y\right)^2\)
\(=\left[3x-9y+10x+5y\right]\left[3x-9y-\left(10x+5y\right)\right]\)
\(=\left(13x-4y\right)\left(-7x-14y\right)\)
\(=-7\left(x+2y\right)\left(13x-4y\right)\)
9(x - 3y)² - 25(2x + y)²
= 3².(x - 3y)² - 5².(2x + y)²
= (3x - 9y)² - (10x + 5y)²
= (3x - 9y - 10x - 5y)(3x - 9y + 10x + 5y)
= (-7x - 14y)(13x - 4y)
= -7(x + 2y)(13x - 4y)
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Lời giải:
Gọi số cần tìm là $\overline{ab}$. Ta có:
$\overline{ab2}=405+\overline{ab}$
$\overline{ab}\times 10+2=405+\overline{ab}$
$\overline{ab}\times 10-\overline{ab}=405-2$
$\overline{ab}\times 9=403$
$\overline{ab}=403:9$ không phải số tự nhiên.
Đề có vẻ sai. Bạn xem lại.
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A D B C I
a/
Ta có
DC=AD+BC (gt)
CI=BC (gt)
=> DC=AD+CI
Ta có
DC=DI+CI
=> AD=DI => tg ADI cân tại D \(\Rightarrow\widehat{DAI}=\widehat{DIA}\)
Mà \(\widehat{DAI}=\widehat{BAI}\)
\(\Rightarrow\widehat{DIA}=\widehat{BAI}\) Mà 2 góc này ở vị trí so le trong
=> AB//CD => ABCD là hình thang
b/
Ta có
CI=BC (gt) => tg BCI cân tại C \(\Rightarrow\widehat{CBI}=\widehat{CIB}\)
Ta có
AB//CD \(\Rightarrow\widehat{ABI}=\widehat{CIB}\) (góc so le trong)
\(\Rightarrow\widehat{CBI}=\widehat{ABI}\) => BI là phân giác của góc B
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a) Ư(12²) = Ư(144) = {1; 2; 3; 4; 6; 8; 9; 12; 16; 18; 24; 32; 48; 72; 144}
b) Ư(18²) = Ư(324) = {1; 2; 3; 4; 6; 9; 12; 18; 27; 36; 54; 81; 108; 162; 324}
c) Ư(24²) = Ư(576) = {1; 2; 3; 4; 6; 8; 9; 12; 16; 18; 24; 32; 36; 48; 64; 72; 96; 144; 192; 288; 576}
d) Ư(32²) = Ư(1024) = {1; 2; 4; 8; 16; 32; 64; 128; 256; 512; 1024}
a) x² + 4x + 4 = (x + 2)²
b) 4x² - 4x + 1 = (2x - 1)²
c) 2x - 1 - x²
= -(x² - 2x + 1)
= -(x - 1)²
d) x² + x + 1/4
= x² + 2.x.1/2 + (1/2)²
= (x + 1/2)²
e) 9 - x²
= 3² - x²
= (3 - x)(3 + x)
g) (x + 5)² - 4x²
= (x + 5)² - (2x)²
= (x + 5 - 2x)(x + 5 + 2x)
= (5 - x)(3x + 5)
h) (x + 1)² - (2x - 1)²
= (x + 1 - 2x + 1)(x + 1 + 2x - 1)
= (2 - x).3x
= 3x(2 - x)
i) Sửa đề: x²y² - 4xy + 4
= (xy)² - 2.xy.2 + 2²
= (xy - 2)²
k) y² - (x² - 2x + 1)
= y² - (x - 1)²
= (y - x + 1)(y + x - 1)
l) x³ + 6x² + 12x + 8
= x³ + 3.x².2 + 3.x.2² + 2³
= (x + 2)³
m) 8x³ - 12x²y + 6xy² - y³
= (2x)³ - 3.(2x)².y + 3.2x.y² - y³
= (2x - y)³