tính bằng cách hợp lí
\(1+8.\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right)-9^8\)
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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 \(AE=15cm\) !!!
I'll let you draw the figurr by yourself.
a) There are 3 trapezoid: BDEC, BDIC and BIEC.
b) Because \(DE//BC\), \(\widehat{DIB}=\widehat{IBC}\) (2 staggered angles)
We also have \(\widehat{IBC}=\widehat{IBD}\) (because BI is the bisector of \(\widehat{ABC}\)). Thus, IBD is an isosceles triangle, which means \(ID=BD\)
Similarly, we have \(IE=CE\). Form these, we have \(ID+IE=BD+CE\) or \(DE=BD+CE\)
And that's what we have to prove!
a) Xét tam giác HBA và tam giác ABC :
góc B chung
góc BHA = góc BAC (= 90 độ)
=> Tam giác HBA đồng dạng với tam giác ABC
b) Áp dụng định lí pytago vào tam giác vuông ABC, ta được :
AB^2 + AC^2 = BC^2
=> 15^2 + 20^2 = BC^2
=> BC^2 = 625
=> BC = 25 (cm)
Vì tam giác HBA đồng dạng với tam giác ABC
=> AH/AC = AB/BC
=> AH/20 = 15/25
=> AH = 20.15/25 = 12 (cm)
No one but you draw the figure.
a) Consider the right triangle ABC, which has \(\widehat{A}=90^o\), we have \(\widehat{B}+\widehat{C}=90^o\Leftrightarrow\widehat{C}=90^o-\widehat{B}\) (1)
On the other hand, the triangle ABC has the height AH, therefore, triangle HBA is also a right triangle \(\left(\widehat{AHB}=90^o\right)\)
Thus, we have \(\widehat{BAH}+\widehat{B}=90^o\Leftrightarrow\widehat{BAH}=90^o-\widehat{B}\) (2)
From (1) and (2), we get \(\widehat{HAB}=\widehat{C}\)
Consider the 2 triangles HAB and ABC, both of these triangles are right triangles, also, \(\widehat{HAB}=\widehat{C}\). Therefore, \(\Delta HAB~\Delta ABC\left(a.a\right)\)
b) Consider the right triangle ABC \(\left(\widehat{A}=90^o\right)\). According to the Pytagorean theorem, we have \(BC^2=AB^2+AC^2\Leftrightarrow BC=\sqrt{AB^2+AC^2}\)
Because \(AB=15cm;AC=20cm\), \(BC=\sqrt{15^2+20^2}=25\left(cm\right)\)
Triangle ABC rights at A, so \(S_{ABC}=\dfrac{1}{2}AB.AC\) (3)
Also, triangle ABC has the height AH, so \(S_{ABC}=\dfrac{1}{2}AH.BC\)(4)
From (3) and (4), we have \(\dfrac{1}{2}AH.BC=\dfrac{1}{2}AB.AC\left(=S_{ABC}\right)\Leftrightarrow AH.BC=AB.AC\)\(\Leftrightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{15.20}{25}=12\left(cm\right)\)
So, \(BC=25cm\) and \(AH=12cm\)
c) What is the question? I can't see it.
a) Ta có: \(Q=12x-4x^2-11=-\left(4x^2-12x+9\right)-2=-\left(2x-3\right)^2-2< 0\)
\(\Rightarrow\)ĐPCM
b) Ta có: \(\left(a^3+b^3\right)\left(a^2+b^2\right)=a^5+b^5+a^3b^2+a^2b^3=a^5+b^5+a^2b^2\left(a+b\right)\)
Mà \(ab=1\Rightarrow\left(a^3+b^3\right)\left(a^2+b^2\right)=a^5+b^5+\left(a+b\right)\)
\(\Rightarrow a^5+b^5=\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\)
a) We have \(Q=12x-4x^2-11=-\left(4x^2-12x+9\right)-2=-\left(2x-3\right)^2-2\)Because \(-\left(2x-3\right)^2\le0\); \(-\left(2x-3\right)^2-2\le-2< 0\Leftrightarrow Q< 0\)
And that's the thing we have to prove.
b) Just expand the polynomial on the right side of the equality:
We have \(R=\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\) \(=a^5+b^5+a^3b^2+a^2b^3-\left(a+b\right)\)\(=a^5+b^5+a^2b^2\left(a+b\right)-\left(a+b\right)\)
On the other hand, \(ab=1\Leftrightarrow a^2b^2=1\)
Therefore, \(R=a^5+b^5+\left(a+b\right)-\left(a+b\right)=a^5+b^5=L\)
Thus, the quality was proved.
a/ \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)\(\Rightarrowđpcm\)
b/ \(x^7+x^5+1=x^7+x^6+x^5-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\left(đpcm\right)\)
\(VT=x^3+y^3+z^3=\left(x+y\right)^3+z^3-3x^2y-3xy^2=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3x^2y-3xy^2-3xyz+3xyz=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)+3xyz=3xyz+\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=3xyz+\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=VP\left(DPCM\right)\)
gọi cạnh sân hình vuông là x đk x > 0
diện tích sân hình vuông là x2
chiều dài sân hình chữ nhật là x + 3
chiều rộng là x - 2
diện tích sân hình chữ nhật là (x+3)(x -2)
theo bài ra ta có x2 = (x+3)(x -2)
x2 = x2 - 2x + 3x - 6
x - 6 = 0
x = 6
vậy cạnh sân hình vuông là 6 m
Gọi cạnh hình vuông là \(x\) (\(x>0\))
Diện tích sân hình vuông là : \(x^2\)
Chiều dài sân hình chữ nhật là : \(x+3\) ; chiều rộng : \(x-2\)
Diện tích sân hình chữ nhật là : \(\left(x+3\right)\left(x-2\right)\)
Ta có : \(x^2=\left(x+3\right)\left(x-2\right)\)
\(\Leftrightarrow x^2=x^2-2x+3x-6\)
\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)
Vậy cạnh sân hình vuông là \(6m\)
a/
Xét tg ABE và tg CDF có
BE=FD (gt)
AB=CD (cạnh đối hbh)
\(\widehat{ABE}=\widehat{CDF}\) (góc so le trong)
=> tg ABE = tg CDF (c.g.c) => AE=CF (1)
Chứng minh tương tự ta cũng có tg ADF = tg BCE
=> AF=CE (2)
Từ (1) (2) => AECF là hbh (Tứ giác có các cặp cạnh đối bằng nhau từng đôi 1)
b/
Xét tg BCF có
BE=EF
EM//CF
=> MB=MC (trong tg đường thẳng // với 1 cạnh và đi qua trung điểm 1 cạnh thì đi qua trung điểm cạnh còn lại)
C/m tương tự khi xét tg CDE => NC=ND
c/
Ta có
\(BE=EF=FD=\dfrac{BD}{3}\Rightarrow BD=3.FE\)
Xét tg BCD có
MB=MC; NC=ND => MN là đường trung bình của tg BCD
\(\Rightarrow MN=\dfrac{BD}{2}=\dfrac{3.EF}{2}\)
\(1+8.\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^2-1\right).\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^8-1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^8\right)^2-1-9^8=\left(3^2\right)^8-9^8=9^8-9^8=0\)