Em hãy tính nhanh các bài toán sau đây:
\(64^2+128.36+36^2\)
\(\left(12^{3+1}\right).\left(12^3-1\right)-3^6.4^6\)
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mik ko chép lại đề bài nha
a) = (123)2- 12- (36. 46)
= (126-1)- (3.4)6
= 126-1-126
= -1
a) \(5x^2-2x\left(3x+\frac{3}{2}\right)=-x^2-3x=-x\left(x+3\right)=-3\left(3+3\right)=-18\)
b) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2=3\left(x^2-\frac{4}{5}y^2\right)\)
\(=3\left(4^2-\frac{4}{5}.5^2\right)=3.\left(-4\right)=-12\)
c) \(\left(x-2\right)^2-\left(x+7\right)\left(x-7\right)=x^2-4x+4-x^2+49=-4x+53=-4.3+53=41\)
d) \(x^2+12x+36=\left(x+6\right)^2=\left(64+6\right)^2=70^2=4900\)
e) \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=x^2-6x+9-x^2+16=-6x+25=-6\left(-1\right)+25\)
= 31
f) \(\left(3x+2y\right)^2-4y\left(3x+y\right)=9x^2+12xy+4y^2-12xy-4y^2=9x^2=9\left(-\frac{1}{3}\right)^2=1\)
a) Đặt: x = a- b; y = b - c ; z = c- a
Ta có: x + y + z = 0
=> \(A=x^3+y^3+z^3=3xyz+\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=3xyz\)
=> \(A=3xyz=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
b) Đặt: \(a=x^2-2x\)
Ta có: \(B=a\left(a-1\right)-6=a^2-a-6=\left(a+2\right)\left(a-3\right)=\left(x^2-2x+2\right)\left(x^2-2x-3\right)\)
\(=\left(x^2-2x+2\right)\left(x+1\right)\left(x-3\right)\)
d) \(D=4\left(x^2+2x-8\right)\left(x^2+7x-8\right)+25x^2\)
Đặt: \(x^2-8=t\)
Ta có: \(D=4\left(t+2x\right)\left(t+7x\right)+25x^2\)
\(=4t^2+36xt+81x^2=\left(2t+9x\right)^2\)
\(=\left(2x^2+9x-16\right)^2\)
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
Ta có : 642 + 128.36 + 362
= 642 + 2.64.36 + 362
= (64 + 36)2
= 1002
= 10000
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Ta có : 642 + 128.36 + 362
= 642 + 2.64.36 + 362
= (64 + 36)2
= 1002
= 10000