Toán

a: HE vuông góc AC

AB vuông góc AC

=>HE//AB

b: Xét ΔCAH vuông tại H và ΔCBA vuông tại A có

góc ACH chung

=>ΔCAH đồng dạng với ΔCBA

c: Xét ΔKEH và ΔKBA có

góc KEH=góc KBA

góc EKH=góc BKA

=>ΔKEH đồng dạng với ΔKBA

=>KE/KB=KH/KA

=>EK/EB=HK/HA

Xét ΔEAB có MK//AB

nên MK/AB=EK/EB

Xét ΔHAB có KN//AB

nên KN/AB=HK/HA

=>MK/AB=KN/AB

=>MK=KN

2x4 chia hết cho 4 khi x=0; x=2; x=4; x=6; x=8 

24x chia hết cho 5 khi x=0; x=5

Tìm x:1/3 : x + 2/3 : x = 4

Diện tích đáy ABC của hình lăng trụ đứng là:

\(25:5=5\left(cm^2\right)\)

Độ dài của BC là:

\(S_{ABC}=\dfrac{1}{2}.AB.BC\)

\(\Rightarrow BC=\dfrac{2.S_{ABC}}{AB}=\dfrac{2.5}{2}=5\left(cm\right)\)

Vậy \(BC=5\left(cm\right)\)

a) \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x+2\right)\left(2-x\right)=1\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(x-2\right)=1\)

\(\Leftrightarrow\left(x^3-3^3\right)-x\left(x+2\right)\left(x-2\right)=1\)

\(\Leftrightarrow x^3-27-x\left(x^2-2^2\right)=1\)

\(\Leftrightarrow x^3-27-x^3+4x=1\)

\(\Leftrightarrow4x-27=1\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

b) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow\left[\left(x+1\right)-\left(x-1\right)\right]\left[\left(x+1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x-1\right)^2\right]-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left[\left(x+1\right)^2+\left(x^2-1\right)+\left(x-1\right)^2\right]-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left[\left(x^2+2x+1\right)+\left(x^2-1\right)+\left(x^2-2x+1\right)\right]-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left(3x^2+1\right)-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow6x^2+2-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=-\dfrac{6}{12}=-\dfrac{1}{2}\)

\(a,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\\ \Leftrightarrow x^3-3x^2+3x^2-9x+9x-27+\left(x^2+2x\right)\left(2-x\right)-1=0\\ \Leftrightarrow x^3-3x^2+3x^2-9x+9x-27+2x^2-x^3+4x-2x^2-1=0\\ \Leftrightarrow x^3-x^3-3x^2+3x^2+2x^2-2x^2-9x+9x+4x=1+27\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)

\(b,\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\\ \Leftrightarrow\left[\left(x+1\right)-\left(x-1\right)\right].\left[\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]-6\left(x^2-2x+1\right)=-10\\ \Leftrightarrow2.\left[\left(x^2+2x+1\right)+\left(x^2-1\right)+\left(x^2-2x+1\right)\right]-6x^2+12x-6=-10\\ \Leftrightarrow2.\left(3x^2+1\right)-6x^2+12x-6=-10\\ \Leftrightarrow6x^2-6x^2+12x=-10+6-2\\ \Leftrightarrow12x=-6\\ \Leftrightarrow x=-\dfrac{1}{2}\)

\(3\cdot\left(-2\right)^2+4\cdot\left(-5\right)+20\)

\(=3\cdot4+4\cdot\left(-5\right)+4\cdot5\)

\(=4\cdot\left[3+\left(-5\right)+5\right]\)

\(=4\cdot3\)

\(=12\)

a) \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(=\left(2x+3\right)\left[\left(2x\right)^2-2x\cdot3+3^2\right]-2\left(4x^3-1\right)\)

\(=\left[\left(2x\right)^3+3^3\right]-2\left(4x^3-1\right)\)

\(=\left(8x^3+27\right)-8x^3+2\)

\(=8x^3+27-8x^3+2\)

\(=29\)

Vậy: ....

c) \(2\left(x^3+y^3\right)-3\left(x^3+y^3\right)\)

\(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3x^2-3y^2\)

\(=2\left(x^2-xy+y^2\right)\cdot1-3x^2-3y^2\)

\(=2x^2-2xy+2y^2-3x^2-3y^2\)

\(=-x^2-2xy-y^2\)
\(=-\left(x^2+2xy+y^2\right)\)

\(=-\left(x+y\right)^2\)

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

Vậy: ...

Bạn gõ đề ở khung \(\Sigma\) cho đề rõ hơn nhé !

a: \(B=\dfrac{x\left(1-x\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^2}{1-x}+x\right)\left(\dfrac{1+x^2}{1+x}-x\right)\right]\)

\(=\dfrac{x\left(x-1\right)^2}{x^2+1}:\left[\dfrac{1-x^2+x-x^2}{1-x}\cdot\dfrac{1+x^2-x-x^2}{1+x}\right]\)

\(=\dfrac{x\left(x-1\right)^2}{x^2+1}\cdot\dfrac{\left(1-x\right)\left(1+x\right)}{\left(-2x^2+x+1\right)\left(-x+1\right)}\)

\(=\dfrac{x\left(x-1\right)^2}{x^2+1}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{-\left(x-1\right)\left(2x^2-x-1\right)}\)

\(=\dfrac{-x\left(x-1\right)^2}{x^2+1}\cdot\dfrac{x+1}{2x^2-2x+x-1}\)

\(=\dfrac{-x\left(x-1\right)^2}{x^2+1}\cdot\dfrac{x+1}{\left(x-1\right)\left(2x+1\right)}\)

\(=\dfrac{-x\left(x-1\right)\left(x+1\right)}{\left(2x+1\right)\left(x^2+1\right)}\)

b: Đề này sai rồi bạn ,lỡ x=2 thì nó nhỏ hơn 0 á bạn

=x^2+16x+64+2(x^2+6x-16)+x^2-4x+4

=2x^2+12x+68+2x^2+12x-32

=4x^2+24x+36

\(\left(x+8\right)\left(x+8\right)-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)

\(=\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)

\(=\left[\left(x+8\right)-\left(x-2\right)\right]^2\)

\(=\left(x+8-x+2\right)^2\)

\(=10^2\)

\(=100\)