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a) \(\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)
\(=\left(2x+y\right)^2-\left(2x\right)^2+y^2+xy-y^2\)
\(=\left(2x+y+2x\right)\left(2x+y-2x\right)+xy\)
\(=\left(4x+y\right)y+xy\)
\(=\left[4\left(-2\right)+3\right].3+\left(-2\right).3\)
\(=\left(-8+3\right).3+1\)
\(=-15+1\)
\(=-14\)
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\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)=\left(a-3b-a-3b\right)\left(a-3b+a+3b\right)-\left(ab-2a-b+2\right)=\left(-6b\right).2a-ab+2a+b-2=2a+b-13ab-2\)
Thay \(a=\dfrac{1}{2};b=-3\) vào N ta được: \(N=2a+b-13ab-2=2.\dfrac{1}{2}-3-13.\dfrac{1}{2}.\left(-3\right)-2=\dfrac{31}{2}\)
Ta có: \(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)
\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)-1-3-2\)
\(=\dfrac{27}{2}\)
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a) N = (a - 3b)2 - (a + 3b)2 - (a - 1)(b - 2)
= [a - 3b + (a + 3b)][a - 3b - (a + 3b)] - [a(b - 2) - 1(b - 2)]
= (a - 3b + a + 3b)(a - 3b - a - 3b) - (ab - 2a - b + 2)
= 2a.(-6b) - ab + 2a + b - 2
= -12ab - ab + 2a + b - 2
= -13ab + 2a + b - 2
Thay a = \(\frac{1}{2}\)và b = -3 vào biểu thức ta có :
N = -13ab + 2a + b - 2 = \(\left(-13\right)\cdot\frac{1}{2}\cdot\left(-3\right)+2\cdot\frac{1}{2}+\left(-3\right)-2=\frac{31}{2}\)
b) P = (2x - 3)(2x + 3) - (2x + 1)2
= (2x)2 - 32 - [(2x)2 + 2.2x.1 + 12 ]
= 4x2 - 9 - (4x2 + 4x + 1)
= 4x2 - 9 - 4x2 + 4x + 1
= (4x2 - 4x2) + (-9 +1) + 4x
= -8 + 4x
Thay x = -2005 vào biểu thức ta có :
P = -8 + 4x = -8 + 4.(-2005) = -8028
c) Q = (y - 3)(y + 3)(y2 + 9) - (y2 + 2)(y2 - 2)
= (y2 - 9)(y2 + 9) - (y2 + 2)(y2 - 2)
= (y2 - 81) - (y2 - 4)
= y2 - 81 - y2 + 4 = -77
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3. Câu hỏi của Hoàng Đức Thịnh - Toán lớp 8 - Học toán với OnlineMath
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a) Ta có: \(\dfrac{3a^2-10a+3}{2\left(a-3\right)}\)
\(=\dfrac{3a^2-9a-a+3}{2\left(a-3\right)}\)
\(=\dfrac{3a\left(a-3\right)-\left(a-3\right)}{2\left(a-3\right)}\)
\(=\dfrac{\left(a-3\right)\left(3a-1\right)}{2\left(a-3\right)}\)
\(=\dfrac{3a-1}{2}\)
\(=\dfrac{3}{2}a-\dfrac{1}{2}\)(đpcm)
b) Ta có: \(\dfrac{b^2+3b+9}{b^3-27}\)\(=\dfrac{b^2+3b+9}{\left(b-3\right)\left(b^2+3b+9\right)}\)
\(=\dfrac{1}{b-3}\)
\(=\dfrac{b-2}{\left(b-3\right)\left(b-2\right)}\)
\(=\dfrac{b-2}{b^2-5b+6}\)(đpcm)
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2: \(N=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)
\(=\dfrac{39}{2}+1-3-2=\dfrac{39}{2}-4=\dfrac{31}{2}\)
3: \(P=4x^2-25-4x^2-4x-1=-4x-26\)
=-8020-26=-8046
4: \(Q=\left(y^2-9\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=y^4-81-y^4+4=-77\)
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a) \(\frac{a-1}{2}=\frac{b-2}{3}=\frac{c-3}{4}\Leftrightarrow\frac{2a-2}{4}=\frac{3b-6}{9}=\frac{c-3}{4}\)
Áp dụng t/c dãy tỉ số bằng nhau : \(\frac{2a-2}{4}=\frac{3b-6}{9}=\frac{c-3}{4}=\frac{2a+3b-c-2-6+3}{4+9-4}=\frac{45}{9}=5\)
Suy ra : \(\begin{cases}a=11\\b=17\\c=23\end{cases}\)
\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)
\(=\left(a-3b+a+3b\right)\left(a-3b-a-3b\right)-\left(ab-2a-b+2\right)\)
\(=2a.\left(-6b\right)-ab+2a+b-2\)
\(=-12ab-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
Thay a=1/2,b=-3 vào N ta được:
\(N=-13\cdot\frac{1}{2}.\left(-3\right)+2\cdot\frac{1}{2}-3-2=\frac{39}{2}+\frac{2}{2}-5=\frac{41}{2}-5=\frac{31}{2}\)
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