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4)
a,
\(34^2+66^2+68\cdot66\\ =34^2+68\cdot66+66^2\\=34^2+2\cdot34\cdot68+66^2\\ =\left(34+66\right)^2\\ =100^2 =10000\)
b,
\(74^2+24^2-48\cdot74\\ =74^2-48\cdot74+24^2\\ =74^2-2\cdot24\cdot74+24^2\\ =\left(74-24\right)^2\\ =50^2=2500\)
c,
\(729^2-728^2\\ =\left(729+728\right)\left(729-728\right)\\ =1457\cdot1\\ =1457\)
d,
\(1001^2-1\\ =1001^2-1^2\\ =\left(1001+1\right)\left(1001-1\right)\\ =1002\cdot1000\\ =1002000\)
5)
a,
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =1\cdot\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2^8-1\right)\left(2^8+1\right)\\ =2^{16}-1\)
b,
\(7\cdot\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^3-1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^6-1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^{12}-1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^{24}-1\right)\left(2^{24}+1\right)\\ =2^{48}-1\)
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\(2\sqrt{27}-\sqrt{\dfrac{16}{3}}-\sqrt{48}-\sqrt{8\dfrac{1}{3}}\)
\(=6\sqrt{3}-4\sqrt{\dfrac{1}{3}}-4\sqrt{3}-5\sqrt{\dfrac{1}{3}}\)
\(=2\sqrt{3}-9\sqrt{\dfrac{1}{3}}\)
\(=2\sqrt{3}-3\sqrt{9\cdot\dfrac{1}{3}}\)
\(=2\sqrt{3}-3\sqrt{3}\)
\(=-\sqrt{3}\)
________________________
\(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\)
\(=\left(5\sqrt{5}-2\sqrt{3}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+3\sqrt{3}\right)\)
\(=\left(3\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}+2\sqrt{3}\right)\)
\(=\left(3\sqrt{5}\right)^2-\left(2\sqrt{3}\right)^2\)
\(=15-12\)
\(=3\)
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Bài 13:
a) \(501^2\)
\(=\left(500+1\right)^2\)
\(=500^2+2\cdot500\cdot1+1^2\)
\(=250000+1000+1\)
\(=251001\)
b) \(88^2+24\cdot88+12^2\)
\(=88^2+2\cdot12\cdot88+12^2\)
\(=\left(88+12\right)^2\)
\(=100^2\)
\(=10000\)
c) \(52\cdot48\)
\(=\left(50+2\right)\left(50-2\right)\)
\(=50^2-2^2\)
\(=2500-4\)
\(=2496\)
Bài 14:
a) \(P=\left(2x-1\right)\left(4x^2+2x+1\right)+\left(x+1\right)\left(x^2-x+1\right)\)
\(P=\left(2x\right)^3-1+x^3+1\)
\(P=8x^3+x^3\)
\(P=9x^3\)
b) \(Q=\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x+y\right)\left(x^2-xy+y^2\right)+2y^3\)
\(Q=x^3-y^3-x^3-y^3+2y^3\)
\(Q=-2y^3+2y^3\)
\(Q=0\)
RÚT GỌN BIỂU THỨC
a) 32(x+2)(x-2)-1/2(6-8x)2 -48
b) (x+9)(x2+27)-(x+3)3
c) (6x+1)2(6x-1)2-2(1+6x)(6x-1)
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a: \(=6\sqrt{2}-12\sqrt{3}-10\sqrt{2}+12\sqrt{3}=-4\sqrt{2}\)
b: \(=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\sqrt{4-3}=1\)
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\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v