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a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)
\(\Leftrightarrow3x-9=90-5+10x\)
\(\Leftrightarrow3x-9=10x+85\)
\(\Leftrightarrow3x-10x=85+9\)
\(\Leftrightarrow-7x=94\)
hay \(x=-\dfrac{94}{7}\)
Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)
b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)
\(\Leftrightarrow6x-4-60=9-6x-42\)
\(\Leftrightarrow6x-64=-6x-33\)
\(\Leftrightarrow6x+6x=-33+64\)
\(\Leftrightarrow12x=31\)
hay \(x=\dfrac{31}{12}\)
Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)
c) Ta có: \(3\left(x-1\right)+3=5x\)
\(\Leftrightarrow3x-3+3=5x\)
\(\Leftrightarrow3x-5x=0\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: S={0}
d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
nên x+101=0
hay x=-101
Vậy: S={-101}
a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)
Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt
b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)
Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt
c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)
Vậy x = 3 là nghiệm của pt
d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)
Vậy x = -101 là nghiệm của pt
e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
1:
a: =7/5(40+1/4-25-1/4)-1/2021
=21-1/2021=42440/2021
b: =5/9*9-1*16/25=5-16/25=109/25
2: \(\left(\dfrac{7}{a+7}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right):\dfrac{a+1}{2}\)
\(=\dfrac{7a-49+a^2+49-7a-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}\)
\(=\dfrac{a^2-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}=\dfrac{2}{a+1}\)
3: \(=\dfrac{x^4-4x^2+4x^2}{x^2-4}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x^2-4\right)}\cdot\dfrac{x^2-4}{x-2}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x-2\right)}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x^2-4\right)+\left(2-3x\right)\left(x-4\right)}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-4x+2x-8-3x^2+12x}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-3x^2+10x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-x^2-2x^2+2x+8x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^3\left(x-1\right)\left(x^2-2x+8\right)}{\left(x-2\right)^2\cdot\left(x+2\right)\left(x-4\right)}\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
Coi như tất cả các biểu thức cần tính đạo hàm đều xác định.
1.
\(y'=2sin\sqrt{4x+3}.\left(sin\sqrt{4x+3}\right)'=2sin\sqrt{4x+3}.cos\sqrt{4x+3}.\left(\sqrt{4x+3}\right)'\)
\(=sin\left(2\sqrt{4x+3}\right).\dfrac{4}{2\sqrt{4x+3}}=\dfrac{2sin\left(2\sqrt{4x+3}\right)}{\sqrt{4x+3}}\)
2.
\(y'=3x^3+\dfrac{17}{x\sqrt{x}}\)
3.
\(y'=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\left(\dfrac{sin4x}{cos\left(x^2+2\right)}\right)'\)
\(=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\dfrac{4cos4x.cos\left(x^2+2\right)+2x.sin4x.sin\left(x^2+2\right)}{cos^2\left(x^2+2\right)}\)
4.
\(y'=-\dfrac{\left(\sqrt{sin^2\left(6-x\right)+4x}\right)'}{sin^2\left(6-x\right)+4x}=-\dfrac{\left[sin^2\left(6-x\right)+4x\right]'}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
\(=-\dfrac{2sin\left(6-x\right).\left[sin\left(6-x\right)\right]'+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}=-\dfrac{-2sin\left(6-x\right).cos\left(6-x\right)+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
\(=\dfrac{sin\left(12-2x\right)-4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
5.
\(y'=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).\left[sin\left(\dfrac{2x-1}{4-x}\right)\right]'\)
\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).cos\left(\dfrac{2x-1}{4-x}\right).\left(\dfrac{2x-1}{4-x}\right)'\)
\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+x.sin\left(\dfrac{4x-2}{4-x}\right).\dfrac{7}{\left(4-x\right)^2}\)
Lời giải:
a.
$\frac{2}{3}x-\frac{7}{6}=\frac{12}{7}-\frac{1}{2}=\frac{17}{14}$
$\frac{2}{3}x=\frac{17}{14}+\frac{7}{6}=\frac{50}{21}$
$x=\frac{50}{21}: \frac{2}{3}=\frac{25}{7}$
b.
$(1\frac{1}{2}+\frac{5}{3}-\frac{1}{6}):x=\frac{3}{4}-\frac{1}{2}$
$3:x=\frac{1}{4}$
$x=3: \frac{1}{4}=12$
a: x=4/27-2/3=4/27-18/27=-14/27
b: =>3/4x-1/4x=1/6+7/3
=>1/2x=1/6+14/6=5/2
hay x=5
c: =>13/10x=7/2+5/2=6
=>x=13/10:6=13/60
d: (3x+2)(-2/5x-7)=0
=>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-35/2
Bài 3
a,26/100+0,009+41/100+0,24
0,26+0,09+0,41+0,24
(0,26+0,24)+(0,09+0,41)
0,5+0,5
=1
b,9+1/4+6+2/7+7+3/5+8+2/3+2/5+1/3+5/7+3/4
(9+6+7+8)+(2/7+5/7)+(1/4+3/4)+(3/5+2/5)+(2/3+1/3)
30+1+1+1+1
=34
Bài 4,5 khó quá mik ko bít lamf^^))
Bài 4: a, \(\dfrac{2008}{2009}\) < 1; \(\dfrac{10}{9}\) > 1
\(\dfrac{2008}{2009}\) < \(\dfrac{10}{9}\)
b, \(\dfrac{1}{a+1}\) và \(\dfrac{1}{a-1}\)
Ta có: a + 1 > a - 1 ⇒ \(\dfrac{1}{a+1}\) < \(\dfrac{1}{a-1}\)
`(-2)/7 xx x = 6/49`
`x =6/49 : (-2)/7`
`x = 6/49 xx (-7)/2`
`x=(-3)/7`
____________________
`2/3x - 1/4 = -3 1/2`
`2/3x - 1/4 = (-7)/2`
`2/3x = (-7)/2 + 1/4`
`2/3x= (-13)/4`
`x = (-13)/4 : 2/3`
`x=(-13)/4 xx 3/2`
`x=(-39)/8`
__________________
`2/3 + 5/3 : x = 1/3`
`5/3 : x = 1/3 - 2/3`
`5/3 : x = (-1)/3`
`x = 5/3 : (-1)/3`
`x = 5/3 xx (-3)`
`x= (-15)/3`
`x=-5`
_______________________
`3/4 - ( x . (-1)/2) = 1 2/3`
`3/4 - ( x . (-1)/2) = 5/3`
`x . (-1)/2 = 3/4 - 5/3`
`x .(-1)/2 = (-11)/12`
`x=(-11)/12 : (-1)/2`
`x=(-11)/12 xx (-2)`
`x=11/6`
_________________________
`#H.J`
a. \(-\dfrac{2}{7}\times x=\dfrac{6}{49}\\ x=\dfrac{6}{49}:-\dfrac{2}{7}\\ x=-\dfrac{3}{7}\)
b. \(\dfrac{2}{3}x-\dfrac{1}{4}=-3\dfrac{1}{2}\\ \dfrac{2}{3}x=-3\dfrac{1}{2}+\dfrac{1}{4}\\ \dfrac{2}{3}x=\dfrac{7}{6}+\dfrac{1}{4}\\ \dfrac{2}{3}x=\dfrac{17}{12}\\ x=\dfrac{17}{12}:\dfrac{2}{3}\\ x=\dfrac{17}{8}\)
c. \(\dfrac{2}{3}+\dfrac{5}{3}:x=\dfrac{1}{3}\\ \dfrac{5}{3}:x=\dfrac{1}{3}-\dfrac{2}{3}\\ \dfrac{5}{3}:x=-\dfrac{1}{3}\\ x=\dfrac{5}{3}:-\dfrac{1}{3}\\ x=-5\)
d. \(\dfrac{3}{4}-\left(x.-\dfrac{1}{2}\right)=1\dfrac{2}{3}\\ x.-\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{5}{3}\\ x.-\dfrac{1}{2}=-\dfrac{11}{12}\\ x=-\dfrac{11}{12}:-\dfrac{1}{2}\\ x=\dfrac{11}{6}\)