1 và 3/4 x -5 =3 và 1/3
2/3 x +1/4 =7/12
1/3+2/5 ( x+1) =1
1/4 +1/3 : 3x = -5
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1) 22x + 1 = 32
=> 22x + 1 = 25
=> 2x + 1 = 5
=> 2x = 5 - 1
=> 2x = 4
=> x = 2
(2) 3.x3 - 100 = 275
=> 3x3 = 275 + 100
=> 3x3 = 375
=> x3 = 375 : 3
=> x3 = 125
=> x3 = 53
=> x = 5
(4) (x - 1)3 - 25 = 72
=> (x - 1)3 = 49 + 32
=> (x - 1)3 = 81
(xem lại đề)
5) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{5}=\frac{x-y}{3-5}=\frac{-4}{-2}=2\)
=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.5=10\end{cases}}\)
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6) Ta có: \(\frac{x}{2}=\frac{y}{3}\) => \(\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{4}\) => \(\frac{y}{15}=\frac{z}{12}\)
=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=\frac{-49}{37}\)
=> \(\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}}\) => \(\hept{\begin{cases}x=-\frac{49}{37}\cdot10=\frac{-490}{37}\\y=-\frac{49}{37}\cdot15=-\frac{735}{37}\\z=-\frac{49}{37}\cdot12=-\frac{588}{37}\end{cases}}\)
Vậy ...
mk lm bài mà mk cho là ''khó'' nhất thôi nha
\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và \(x+y+z=-49\)
\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\left(1\right)\)
\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{15}=\frac{z}{12}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)
ADTC dãy tỉ số bằng nhau ta có
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=-\frac{49}{37}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{49}{37}.10=-\frac{490}{37}\\y=-\frac{49}{37}.15=-\frac{735}{37}\\z=-\frac{49}{37}.12=-\frac{588}{37}\end{cases}}}\)
\(1\frac{1}{2}+x=\frac{3}{2}-7\)
<=> \(\frac{3}{2}+x=\frac{-11}{2}\)
<=> \(x=-7\)
\(\frac{1}{4}+\frac{1}{3}:3x=-5\)
<=> \(\frac{1}{3}:3x=\frac{-21}{4}\)
<=> \(3x=\frac{-4}{63}\)
<=> \(x=\frac{4}{189}\)
\(\frac{4}{5}.x=\frac{8}{35}\)
<=> \(x=\frac{2}{7}\)
\(\frac{2}{3x}-\frac{1}{4}=\frac{7}{1}\)
<=> \(\frac{2}{3x}=\frac{29}{4}\)
=> \(8=87x\)
<=> \(x=\frac{8}{87}\)
\(\frac{3}{5x}+\frac{1}{2}=\frac{1}{7}\)
<=> \(\frac{3}{5x}=\frac{-5}{14}\)
<=> \(-25x=42\)
<=> \(x=\frac{-42}{25}\)
\(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16.\frac{2}{3}\right)=0\)
<=> \(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{-32}{3}=0\)
<=> \(\frac{43}{8}+x-\frac{173}{24}=\frac{-32}{3}\)
<=> \(\frac{43}{8}+x=\frac{-83}{24}\)
<=> \(x=\frac{-53}{6}\)
học tốt
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
6: =>x=9/10+1/5=9/10+2/10=11/10
7: =>x=3/8-5/12=36/96-40/96=-1/24
8: =>x=7/6-5/4=14/12-15/12=-1/12
9: =>x=1/35+2/7=1/35+10/35=11/35
10: =>x=2/7-7/10=20/70-49/70=-29/70
a) = 17/19 - 17/19 + 27/35 + 35/35 = 0 + 62/35
b) = 1/3 x 4/5 + 1/3 x6/5 + 1/3 x 2 = 1/3(4/5 + 6/5 + 2) = 1/3 x 4 = = 4/3
c) 4/7 x 2/9 + 4/7 x 7/9 + 2/3 = 4/7 x (2/9 + 7/9) + 2/3 = 4/7 x 1 + 2/3 = 26/21
A) 17/19 - 17/19 + 27/35 + 35/35 = 0 + 62/35
B) 1/3 x 4/5 + 1/3 x 6/5 + 1/3 x 2 = 1/3 x(4/5 + 6/5 x 2 ) = 1/3 x 4 = 4/3
c) TƯƠNG TỰ CÂU A VÀ B
* HOKTOT*
NHA
a)\(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\text{⇔}\dfrac{7}{4}x-5=\dfrac{10}{3}\text{⇔}\dfrac{7}{4}x=\dfrac{25}{3}\text{⇔}x=\dfrac{100}{21}\)
b)\(\dfrac{2}{3}x+\dfrac{1}{4}=\dfrac{7}{12}\text{⇔}\dfrac{2}{3}x=\dfrac{1}{3}\text{⇔}x=\dfrac{1}{2}\)
c)\(\dfrac{1}{3}+\dfrac{2}{5}\left(x+1\right)=1\text{⇔}\dfrac{2}{5}\left(x+1\right)=\dfrac{2}{3}\text{⇔}x+1=\dfrac{5}{3}\text{⇔}x=\dfrac{2}{3}\)
d)\(\dfrac{1}{4}+\dfrac{1}{3}:3x=-5\text{⇔}\dfrac{1}{3}:3x=-\dfrac{21}{4}\text{⇔}\dfrac{1}{9x}=-\dfrac{21}{4}\text{⇔}9x=-\dfrac{4}{21}\text{⇔}x=-\dfrac{4}{189}\)
a, \(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\)
\(\Rightarrow\dfrac{7}{4}x=5+\dfrac{10}{3}=\dfrac{25}{3}\)
\(\Rightarrow x=\dfrac{25}{3}:\dfrac{7}{4}=\dfrac{100}{21}\)
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b, \(PT\Leftrightarrow\dfrac{2}{3}x=\dfrac{7}{12}-\dfrac{1}{4}=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\)
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c, \(PT\Leftrightarrow\dfrac{2}{5}\left(x+1\right)=1-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\Rightarrow x+1=\dfrac{2}{3}:\dfrac{2}{5}=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{3}-1=\dfrac{2}{3}\)
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d, \(PT\Leftrightarrow\dfrac{1}{3}:3x=-5-\dfrac{1}{4}=\dfrac{1}{9}x=-\dfrac{21}{4}\)
\(\Rightarrow x=-\dfrac{189}{4}\)
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