120x2/5+120x3/4+120 x1
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a) `(2x+5)^3-(2x-5)^3-(120x^2+49)`
`=(2x+5-2x+5)[(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)`
`=10(12x^2+25)-(120x^2+49)`
`=120x^2+250-120x^2-49`
`=201`
b) `(4-5x)^2-(3+5x)^2=(4-5x+3+5x)(4-5x-3-5x)=7.(-10x+1)=-70x+7`
Lời giải:
a.
$(2x+5)^3-(2x-5)^3-(120x^2+49)$
$=[(2x+5)-(2x-5)][(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)$
$=10(4x^2+20x+25+4x^2-25+4x^2-20x+25)-(120x^2+49)$
$=10(12x^2+25)-(120x^2+49)=250-49=201$
b.
$(4-5x)^2-(3+5x)^2=[(4-5x)+(3+5x)][(4-5x)-(3+5x)]$
$=7(1-10x)$
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a) A=(4-5x)2-(3+5x)2=(4-5x-3-5x)(4-5x+3+5x)=(-25x+1)1=-25x+1
B=(3x-1)(1+3x)-(3x+1)2=9x2-1-(3x+1)2=9x2-1-(9x2+6x+1)=9x2-1-9x2-6x-1=-6x-2=-2(3x+1)
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\(\left(12\cdot15-x\right)\cdot\dfrac{1}{4}=120\cdot\dfrac{1}{4}\)
\(\Rightarrow\left(180-x\right)\cdot\dfrac{1}{4}=30\)
\(\Rightarrow180-x=30:\dfrac{1}{4}\)
\(\Rightarrow180-x=120\)
\(\Rightarrow x=180-120\)
\(\Rightarrow x=60\)
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(x + 5 ) x \(\frac{19}{13}=57\)
x + 5 = \(57\div\frac{19}{13}=39\)
x = 39 -5 = 34
b) 12 . 15 - x = 120 \(\cdot\frac{1}{4}\div\frac{1}{4}=120\)
180 - x = 120
x = 180 - 120 = 60
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\(1\dfrac{1}{2}x1\dfrac{1}{3}x1\dfrac{1}{4}x1\dfrac{1}{5}x1\dfrac{1}{6}x1\dfrac{1}{7}x1\dfrac{1}{8}x1\dfrac{1}{9}\)
\(=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}x\dfrac{8}{7}x\dfrac{9}{8}x\dfrac{10}{9}\)
\(=x^7.\dfrac{3.4.5.6.7.8.9.10}{2.3.4.5.6.7.8.9}\)
\(=x^7.\dfrac{10}{2}\)
\(=5x^7\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times...\times\dfrac{9}{8}\times\dfrac{10}{9}=\dfrac{10}{2}=5\)
\(120\times\dfrac{2}{5}+120\times\dfrac{3}{4}+120\times1\\ =120\times\left(\dfrac{2}{5}+\dfrac{3}{4}+1\right)\\ =120\times\left(\dfrac{8}{20}+\dfrac{15}{20}+\dfrac{20}{20}\right)\\ =120\times\dfrac{43}{20}\\ =258\)
\(\dfrac{120\times2}{5}hay120\times\dfrac{2}{5}\) hả em?