Tìm số nguyên x:
a. x + 1/5 = 6/15
b. -2x + 4/3 = 12/18
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16
7+x=-16
x=-16-7
x=-23
2) 2x – 35 = 15
2x=15+35
2x=50
x=50:2
x=25
3) 3x + 17 = 12
3x=12-17
3x=-5
x=-5/3
4) (2x – 5) + 17 = 6
2x-5=6-17
2x-5=-11
2x=-11+5
2x=-6
x=-6:2
x=-3
5) 10 – 2(4 – 3x) = -4
2(4-3x)=10-(-4)
2(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
6) - 12 + 3(-x + 7) = -18
3(-x+7)=-18-(-12)
3(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
a. \(=\left(\dfrac{12}{15}+\dfrac{3}{15}\right)+\left(\dfrac{4}{3}+\dfrac{5}{3}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =1+2+1=4\)
b. \(x\times\dfrac{3}{4}=\dfrac{1\times3\times6\times5}{6\times5\times5\times3}\\ x\times\dfrac{3}{4}=\dfrac{1}{5}\\ x=\dfrac{1}{5}:\dfrac{3}{4}\\ x=\dfrac{4}{15}\)
\(a,x-36:18=12-15\\ \Rightarrow x-2=-3\\ \Rightarrow x=-1\\ b,92-\left(17+x\right)=72\\ \Rightarrow17+x=20\\ \Rightarrow x=3\\ c,720:\left[41-\left(2x+5\right)\right]=40\\ \Rightarrow41-\left(2x+5\right)=18\\ \Rightarrow2x+5=23\\ \Rightarrow2x=18\\ \Rightarrow x=9\\ d,\left(x+2\right)^3-23=41\\ \Rightarrow\left(x+2\right)^3=64\\ \Rightarrow\left(x+2\right)^3=4^3\\ \Rightarrow x+2=4\\ \Rightarrow x=2\)
Giup mình với ah.
1- Tính :
A= 5. | x- 5 | - 3x + 1
2 - Tìm các số nguyên x,y ; sao cho :
a) 5/x - y/3 = 1/6 b) 5/x + y/4 = 1/8
3- Tìm giá trị lớn nhất của Q = 27-2x/12-x ( x là số nguyên)
---------------------------------------------------------------------------------------------
1) \(A=5.\left|x-5\right|-3x+1\)
\(A=\left[{}\begin{matrix}5.\left(x-5\right)-3x+1\left(x-5\ge0\right)\\5.\left(5-x\right)-3x+1\left(x-5< 0\right)\end{matrix}\right.\)
\(A=\left[{}\begin{matrix}5x-25-3x+1\left(x\ge5\right)\\25-5x-3x+1\left(x< 5\right)\end{matrix}\right.\)
\(A=\left[{}\begin{matrix}2x-24\left(x\ge5\right)\\26-8x\left(x< 5\right)\end{matrix}\right.\)
3:
\(Q=\dfrac{27-2x}{12-x}=\dfrac{2x-27}{x-12}\)
\(\Leftrightarrow Q=\dfrac{2x-24-3}{x-12}=2-\dfrac{3}{x-12}\)
Để Q lớn nhất thì \(2-\dfrac{3}{x-12}\) lớn nhất
=>\(\dfrac{3}{x-12}\) nhỏ nhất
=>x-12 là số nguyên âm lớn nhất
=>x-12=-1
=>x=11
Vậy: \(Q_{min}=2-\dfrac{3}{11-12}=2+3=5\) khi x=11
Bài 2:
a: \(\dfrac{5}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)
=>\(\dfrac{15-xy}{3x}=\dfrac{1}{6}\)
=>\(15-xy=\dfrac{x}{2}\)
=>\(30-2xy=x\)
=>x+2xy=30
=>x(2y+1)=30
mà x,y nguyên
nên \(\left(x;2y+1\right)\in\left\{\left(30;1\right);\left(-30;-1\right);\left(2;15\right);\left(-2;-15\right);\left(10;3\right);\left(-10;-3\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(30;0\right);\left(-30;-1\right);\left(2;7\right);\left(-2;-8\right);\left(10;1\right);\left(-10;-2\right)\right\}\)
b: \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
=>\(\dfrac{20+xy}{4x}=\dfrac{1}{8}\)
=>\(\dfrac{40+2xy}{8x}=\dfrac{x}{8x}\)
=>40+2xy=x
=>x-2xy=40
=>x(1-2y)=40
mà x,y nguyên
nên \(\left(x;1-2y\right)\in\left\{\left(40;1\right);\left(-40;-1\right);\left(8;5\right);\left(-8;-5\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(40;0\right);\left(-40;1\right);\left(8;-2\right);\left(-8;3\right)\right\}\)
1. Giải:
Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)
\(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)
\(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)
Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.
⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)
Ta có bảng:
2x+1 | 1 | 3 | 7 | 21 |
x | 0 | 1 | 3 | 10 |
TM | TM | TM | TM |
Vậy xϵ\(\left\{0;1;3;10\right\}.\)
2. Giải:
Do (2x-18).(3x+12)=0.
⇒ 2x-18=0 hoặc 3x+12=0.
⇒ 2x =18 3x =-12.
⇒ x =9 x =-4.
Vậy xϵ\(\left\{-4;9\right\}.\)
3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.
S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.
S= 0 + 0 + ... + 0 + 2025.
⇒S= 2025.
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)
\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)
hay x=0
Vậy: x=0
b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)
\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)
hay \(x=\dfrac{-1}{9}\)
Vậy: \(x=\dfrac{-1}{9}\)