C
Bài 1 Tính
a) (3x +y)^2 b) (5-2y)^2
c) (x+2/3) d) (2x-1/2y)^2
e) (3-2x)^3 f) (x^2 + 1/3y).(x^2-1.3y)
mn ơi giúp m với ạ
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Diện tích xung quanh của hình hộp chữ nhật là:
\(2\times4\times\left(5+8\right)=104\left(cm^2\right)\)
Diện tích 2 đáy là:
\(2\times5\times8=80\left(cm^2\right)\)
Diện tích toàn phần của hình hộp là:
\(104+80=184\left(cm^2\right)\)
Thể tích hình hộp là:
\(5\times8\times4=160\left(cm^3\right)\)
Xét \(\triangle ABC\) vuôg tại `A` có: `AD` là đườg p/g
\(=>\dfrac{AB}{AC}=\dfrac{BD}{DC}=\dfrac{3}{4}\)
\(=>AB=\dfrac{3}{4}AC\)
Xét \(\triangle ABC\) vuôg tại `A` có: `AH` là đườg cao
\(=>\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{AH^2}\)
\(=>\dfrac{1}{(\dfrac{3}{4}AC)^2}+\dfrac{1}{AC^2}=\dfac{1}{(\dfrac{12}{5})^2}\)
\(=>AC=4=>AB=\dfrac{3}{4} .4=3\)
Xét \(\triangle ABC\) vuôg tại `A` có: `AB^2+AC^2=BC^2`
`=>3^2+4^2=BC^2=>BC=5`
Chiều rộng hình hộp chữ nhật là : \(6\cdot\dfrac{1}{2}=3\left(cm\right)\)
Chiều cao hình hộp chữ nhật là : \(3\cdot3=9\left(cm\right)\)
\(S_{\text{xq}}\) là : \(\left(6+3\right)\cdot2\cdot9=162\left(cm^2\right)\)
\(S_{\text{tp}}\) là : \(162+2\left(6\cdot3\right)=198\left(cm^2\right)\)
\(V\) là : \(3\cdot6\cdot9=163\left(cm^3\right)\)
Chiều rộng hình HCN là:
6 : 2 = 3 ( cm )
Chiều cao hình HCN là :
3 x 3 = 9 ( cm )
Sxung quanh hình HCN là
( 6+3 ) x 2 x 9 = 162 ( cm2 )
Stoàn phần hình HCN là :
162 + ( 6 x 3 x 2 ) = 198 ( cm2 )
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1\)
\(1+8.\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^2-1\right).\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^8-1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^8\right)^2-1-9^8=\left(3^2\right)^8-9^8=9^8-9^8=0\)
c) This makes no sense. What is quadrilateral ABCE while B, E, C are collinear? May be you mean ABCD but you make some mistake when typing.
ADCE is a parallelogram, therefore, \(CE//AD\) or \(BC//AD\)
Thus, we can easily prove that ABCD is a trapezoid.
Consider the triangle ABE, its height AH is also a median. This means ABE is an isosceles triangle, whose bisector is also AH. Thus, \(\widehat{BAH}=\widehat{EAH}\)
Clearly, we get \(\widehat{BAH}=\widehat{ECA}\left(=90^o-\widehat{B}\right)\), so, \(\widehat{EAH}=\widehat{ECA}\left(=\widehat{BAH}\right)\)
On the other hand, \(EC//AD\Rightarrow\widehat{ECA}=\widehat{CAD}\) (2 staggered angles)
Thus, \(\widehat{EAH}=\widehat{CAD}\left(=\widehat{ECA}\right)\) or \(\widehat{EAH}+\widehat{EAC}=\widehat{CAD}+\widehat{EAC}\) or \(\widehat{CAH}=\widehat{DAE}\)
ADCE is a parallelogram, therefore, \(\widehat{DAE}=\widehat{DCE}\), so, \(\widehat{CAH}=\widehat{DCE}\left(=\widehat{DAE}\right)\) or \(\widehat{CAH}=\widehat{DCB}\)
We also have \(\widehat{CAH}=\widehat{B}\left(=90^o-\widehat{ACB}\right)\), so \(\widehat{B}=\widehat{DCB}\left(=\widehat{CAH}\right)\)
Consider the trapezoid ABCD (AD//BC), it has \(\widehat{B}=\widehat{DCB}\), therefore, ABCD is an isosceles trapezoid.
d) It's pretty easy! Since ABE is an isosceles triangle, \(AE=AB\). Guess what? We've already had \(AB=15cm\). So, simply, we get \(AE=15cm\) !!!
a, \(\left(3x+y\right)^2=9x^2+6xy+y^2\)
b, \(\left(5-2y\right)^2=25-20y+4y^2\)
d, \(\left(2x-\dfrac{1}{2}y\right)^2=4x^2-2.2x\left(\dfrac{1}{2}y\right)+\dfrac{1}{4}y^2=4x^2-2xy+\dfrac{1}{4}y^2\)
e, \(\left(3-2x\right)^3=27-3.9.2x+3.3.4x^2-8x^3=27-54x+36x^2-8x^3\)
f, \(=x^4-\dfrac{1}{9}y^2\)
\(\left(3x+y\right)^2=9x^2+6xy+y^2\\ \left(5-2y\right)^2=25-20y+4y^2\\ \left(x+\dfrac{2}{3}\right)^2=x^2+\dfrac{4}{3}x+\dfrac{4}{9}\\ \left(2x-\dfrac{1}{2}y\right)^2=4x^2-2xy+\dfrac{1}{4}y^2\)