\(- | \dfrac{1}{2} x + 1 | - 15 = 15 \dfrac{1}{4}\)
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a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)
Vậy ...
2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)
Vậy ...
3) PT \(\Rightarrow3x-4-2x+5=3\)
\(\Rightarrow x=2\)
Vậy ...
4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy ...
3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)
\(\Leftrightarrow3x-4-2x+5=3\)
\(\Leftrightarrow x+1=3\)
hay x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
- Thay từng giá trị vào, ta thấy A. \(\dfrac{15}{4}\) thỏa mãn.
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$x:3\frac{1}{15}-\frac{3}{4}=2\frac{1}{4}$
$x:\frac{46}{15}-\frac{3}{4}=\frac{9}{4}$
$x: \frac{46}{15}=\frac{9}{4}+\frac{3}{4}=3$
$x=3\times \frac{46}{15}=\frac{46}{5}$
b. $x\times 3\frac{2}{3}-1\frac{2}{3}=2\frac{1}{3}$
$x\times \frac{11}{3}=1\frac{2}{3}+2\frac{1}{3}=4$
$x=4: \frac{11}{3}=\frac{12}{11}$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(2+4+6+...+98\right)\left(6-6\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\right)\)
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)
- \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) < \(x\) < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)
- \(\dfrac{46}{3}\) < \(x\) < - \(\dfrac{13}{7}\)
\(x\) \(\in\) {-15; -14;-13;..; -2}
a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)
Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)
Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)
Suy ra \(-15\le x\le-2\), x ϵ Z
b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)
Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)
Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)
Suy ra \(-1\le x\le0\), x ϵ Z
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3}{16}\) - (\(x\) - \(\dfrac{5}{4}\)) - ( \(\dfrac{3}{4}\) - \(\dfrac{7}{8}\) - 1) = 2\(\dfrac{1}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{7}{8}\) + 1 = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + ( \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\)) + (\(\dfrac{7}{8}\) + 1) = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\) = \(\dfrac{5}{2}\)
( \(\dfrac{3}{16}\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\)) - \(x\) = \(\dfrac{5}{2}\)
\(\dfrac{41}{16}\) - \(x\) = \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{41}{16}\) - \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{1}{16}\)
2, \(\dfrac{1}{2}\).( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\)) = \(\dfrac{1}{5}\) - \(x\) + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\))
\(\dfrac{1}{2}\).(-\(\dfrac{11}{15}\)) = \(\dfrac{1}{5}\) - \(x\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{5}\)
- \(\dfrac{11}{30}\) = ( \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{15}\)) - \(x\)
- \(\dfrac{11}{30}\) = \(\dfrac{7}{15}\) - \(x\)
\(x\) = \(\dfrac{7}{15}\) + \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)
\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)
\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))
\(\Leftrightarrow x=-36\).
Vậy nghiệm của pt là x = -36.
2) x(x+1)(x+2)(x+3)= 24
⇔ x.(x+3) . (x+2).(x+1) = 24
⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24
Đặt \(x^2\)+ 3x = b
⇒ b . (b+2)= 24
Hay: \(b^2\) +2b = 24
⇔\(b^2\) + 2b + 1 = 25
⇔\(\left(b+1\right)^2\)= 25
+ Xét b+1 = 5 ⇒ b=4 ⇒ \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0
⇒(x-1)(x+4)=0⇒x=1 và x=-4
+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0
⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))2 = - \(\dfrac{15}{4}\) Hay ( \(x^2\) +\(\dfrac{3}{2}\) )2= -\(\dfrac{15}{4}\) (vô lí)
⇒x= 1 và x= 4
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`1,`
`3/16 - (x - 5/4) - (3/4 + (-7)/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-1/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-9/8) = 2 1/2`
`=> 3/16 - x + 19/8 = 2 1/2`
`=> 3/16 - x = 2 1/2 - 19/8`
`=> 3/16 - x =1/8`
`=> x = 3/16 - 1/8`
`=> x = 1/16`
Vậy, `x = 1/16`
`2,`
`1/2* (1/6 - 9/10) = 1/5 - x + (1/15 - (-1)/5)`
`=> 1/2 * (-11/15) = 1/5 - x + 4/15`
`=> -11/30 = x + 1/5 - 4/15`
`=> x + (-1/15) = -11/30`
`=> x = -11/30 + 1/15`
`=> x = -3/10`
Vậy, `x = -3/10.`
\(-\left|\frac{1}{2}x+1\right|-15=15\frac{1}{4}\)
=> \(-\left|\frac{1}{2}x+1\right|=15\frac{1}{4}+15\)
=> \(-\left|\frac{1}{2}x+1\right|=\frac{121}{4}\)
=> x không thỏa mãn
\(-\left|\frac{1}{2}x+1\right|-15=\frac{61}{4}\)
\(-\left|\frac{1}{2}x+1\right|=\frac{121}{4}\)
\(\left|\frac{1}{2}x+1\right|=-\frac{121}{4}\)
Vì \(\left|\frac{1}{2}x+1\right|\ge0\)
Mà \(-\frac{121}{4}< 0\)(ngoặc ''}'' 2 điều lại)
\(\Rightarrow\)Không có giá trị nào của x để thỏa mãn đề bài