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a: Xét ΔABC vuông tại A và ΔHBA vuông tại H có
góc B chung
=>ΔABC đồng dạng với ΔHBA
=>AB/HB=AC/HA
=>AB*HA=HB*AC
b: AH=căn 5^2-3^2=4cm
BI là phân giác
=>HI/HB=IA/AB
=>HI/3=IA/5=(HI+IA)/(3+5)=0,5
=>HI=1,5cm; IA=1,5cm
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a)\(x\left(x-3\right)-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
b)\(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)
\(\Leftrightarrow15x^2-46x+35-15x^2+7x+2-4=0\)
\(\Leftrightarrow33-39x=0\Leftrightarrow33=39x\Leftrightarrow x=\frac{33}{39}\)
a) \(x\left(x-3\right)-2x+6=0\)
\(x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
b) \((3x-5)(5x-7)+(5x+1)(2-3x)=4\)
\(15x^2-46x+35+10x-15x^2+2-3x-4=0\)
\(33-39x=0\)
\(3\left(11-13x\right)=0\)
\(11-13x=0\)
\(13x=11\)
\(x=\frac{11}{13}\)
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c) \(=\left(4x-3\right)^2-\left(9x^2-4\right)\)
\(=16x^2-24x+9-9x^2+4=7x^2-24x+13\)
d) \(=\left(x^2-3x+2\right)\left(x+3\right)-\left(x^3-5x^2\right)\)
\(=x^3+3x^2-3x^2-9x+2x+6-x^3+5x^2\)
\(=5x^2-7x+6\)
c. (4x - 3)(4x - 3) - (3x + 2)(3x - 2)
= (4x - 3)2 - (9x2 - 4)
= 16x2 - 24x + 9 - 9x2 + 4
= 16x2 - 9x2 - 24x + 9 + 4
= 7x2 - 24x + 13
d. (x - 2)(x - 1)(x + 3) - x2(x - 5)
= (x2 - 1 - 2x + 2)(x + 3) - x2(x - 5)
= x3 + 3x2 - x - 3 - 2x2 - 6x + 2x + 6 - x3 + 5
= x3 - x3 + 3x2 - 2x2 - x - 6x + 2x + 6 + 5 - 3
= x2 - 5x + 8
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1: \(=\dfrac{\dfrac{x^2+y^2-2xy}{xy}}{\dfrac{x^2-y^2}{xy}}\)
\(=\dfrac{\left(x-y\right)^2}{xy}:\dfrac{\left(x-y\right)\left(x+y\right)}{xy}=\dfrac{x-y}{x+y}\)
2: \(=\dfrac{x^2-1+x^2}{x\left(x+1\right)}:\dfrac{x^2-x^2+1}{x\left(x+1\right)}\)
\(=\dfrac{2x^2-1}{1}=2x^2-1\)
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a) Diện tích xung quanh của khối rubik là:
\(\dfrac{1}{2}\cdot234\cdot67,5=7897,5\left(mm^2\right)\)
Diện tích mỗi mặt của khối rubik là:
\(7897,5:3=2632,5\left(mm^2\right)\)
Diện tích toàn phần của khối rubik là:
\(2632,5\cdot4=10530\left(mm^2\right)\)
\(---\)
b) Thể tích của khối rubik là:
\(\dfrac{1}{3}\cdot2632,5\cdot63,7=55896,75\left(mm^3\right)\)
Vậy: ...
#\(Toru\)
Trả lời:
Câu 1:
a, \(\left(5x+1\right)\left(2x-3\right)-\left(7x-2\right)\left(x+3\right)+5\)
\(=10x^2-15x+2x-3-\left(7x^2+21x-2x-6\right)+5\)
\(=10x^2-13x-3-7x^2-21x+2x+6+5\)
\(=3x^2-32x+8\)
b, \(\left(3x-2\right)^2+\left(2x+5\right)^2-\left(3x+4\right)\left(3x-4\right)\)
\(=9x^2-12x+4+4x^2+20x+25-9x^2+16\)
\(=4x^2+8x+45\)
c, \(\left(3x+2\right)\left(9x^2-6x+4\right)+\left(5-3x\right)\left(25+15x+9x^2\right)-9\)
\(=27x^3+8+125-27x^3-9\)
\(=124\)
d, \(\left(x-3\right)^3-\left(x+2\right)\left(x-2\right)\left(x-1\right)\)
\(=x^3-9x^2+27x-27-\left(x^2-4\right)\left(x-1\right)\)
\(=x^3-9x^2+27x-27-\left(x^3-x^2-4x+4\right)\)
\(=x^3-9x^2+27x-27-x^3+x^2+4x-4\)
\(=-8x^2+31x-31\)
e, \(\left(x+2y\right)^3-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=x^3+6x^2y+12xy^2+8y^3-\left(8x^3-y^3\right)\)
\(=x^3+6x^2y+12xy^2+8y^3-8x^3+y^3\)
\(=-7x^3+9y^3+6x^2y+12xy^2\)
f, \(\left(x-3y\right)^3-\left(2x-y\right)\left(3x+2y\right)\left(3x-2y\right)\)
\(=x^3-9x^2y+27xy^2-\left(2x-y\right)\left(9x^2-4y^2\right)\)
\(=x^3-9x^2y+27xy^2-27y^3-\left(18x^3-8xy^2-9x^2y+4y^3\right)\)
\(=x^3-9x^2y+27xy^2-27y^3-18x^3+8xy^2+9x^2y-4y^3\)
\(=-17x^3-31y^3+35xy^2\)
Câu 2 : a ) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=-13x=26\Leftrightarrow x=-2\)
b) \(\left(3x+1\right)^2+\left(2x-5\right)^2=13\left(x+2\right)\left(x-2\right)\)
\(\Leftrightarrow9x^2+6x+1+4x^2-20x+25=13\left(x^2-4\right)\)
\(\Leftrightarrow13x^2-14x+26=13x^2-52\Leftrightarrow13x^2-14x+26-13x^2+52=0\)
\(\Leftrightarrow78-14x=0\Leftrightarrow78=14x\Leftrightarrow x=\frac{78}{14}=\frac{39}{7}\)