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10 tháng 10 2021

TL

x ∈ ∅

HT

10 tháng 10 2021

x=4 nhé

28 tháng 11 2023

1) \(A=5.\left|x-5\right|-3x+1\)

\(A=\left[{}\begin{matrix}5.\left(x-5\right)-3x+1\left(x-5\ge0\right)\\5.\left(5-x\right)-3x+1\left(x-5< 0\right)\end{matrix}\right.\)

\(A=\left[{}\begin{matrix}5x-25-3x+1\left(x\ge5\right)\\25-5x-3x+1\left(x< 5\right)\end{matrix}\right.\)

\(A=\left[{}\begin{matrix}2x-24\left(x\ge5\right)\\26-8x\left(x< 5\right)\end{matrix}\right.\)

29 tháng 11 2023

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 tháng 9 2021

\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)

\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)

\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)

1 tháng 9 2021

\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)

\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)

\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)

\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)

17 tháng 5 2016

c)3(2x-1)-5(x-3)+6(3x-4)=24

<=>6x-3-5x-15+18x-24=24

<=>19x-12=24

<=>19x=36

<=>x=\(\frac{36}{19}\)

d)2x(5-3x)+2x(3x-5)-3(x-7)=3

<=>10x-6x2+6x2-10x-3x-21=3

<=>-3(x-7)=3

<=>21-3x=3

<=>-3x=-18

<=>x=6

10 tháng 3 2017

Thêm nữa câu a) Tính: M(x) + N(x)+ P(x)

B) Tính M(x) - N (x) - P(x)

ok rồi giúp mình với nha

9 tháng 3 2022

Ta có \(A\left(x\right)=\dfrac{1}{3}x+1=0\Leftrightarrow x=-1:\dfrac{1}{3}=-3\)

\(B\left(x\right)=-\dfrac{3}{4}x+\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}\left(-\dfrac{3}{4}\right)=4\)

\(C=\left(2x-4\right)\left(x+1\right)=0\Leftrightarrow x=2;x=-1\)

\(D\left(x\right)-4x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)

11 tháng 9 2021

d

3 tháng 4 2017

a/ P(x) = (x - 3)(x + 4)

Ta có: (x - 3)(x + 4) = 0

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

Vậy................................

b/ Q(x) = \(\left(\dfrac{1}{3}x-1\right)\left(2x-\dfrac{3}{5}\right)\)

Ta có: \(\left(\dfrac{1}{3}x-1\right)\left(2x-\dfrac{3}{5}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x-1=0\\2x-\dfrac{3}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=1\Rightarrow x=3\\2x=\dfrac{3}{5}\Rightarrow x=\dfrac{3}{10}\end{matrix}\right.\)

Vậy................................

a: P(x)=2x^5-2x^5+4x^4-3x^4+5=x^4+5

Q(x)=-5x^4+2x^4-x^3+3x^2-10x+2

=-3x^4-x^3+3x^2-10x+2

b: P(x)+Q(x)

=x^4+5-3x^4-x^3+3x^2-10x+2

=-2x^4-x^3+3x^2-10x+7

Q(x)-P(x)

=-3x^4-x^3+3x^2-10x+2-x^4-5

=-4x^4-x^3+3x^2-10x-3

P(x)-Q(x)=-(Q(x)-P(x))

=4x^4+x^3-3x^2+10x+3