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a)
\(\left|x\right|-2\left|x\right|+3\left|x\right|=16+6\left|x\right|-19\)
\(\left|x\right|-2\left|x\right|+3\left|x\right|-6\left|x\right|=16-19\)
\(\left|x\right|.\left(1-2+3-6\right)=-3\)
\(\left|x\right|.\left(-4\right)=-3\)
\(\left|x\right|=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)
b,
2.(|x| - 5) - 15 = 9
\(2.\left(\left|x\right|-5\right)=9+15\)
\(2.\left(\left|x\right|-5\right)=24\)
\(\left|x\right|-5=24:2\)
\(\left|x\right|-5=12\)
\(\left|x\right|=12+5\)
\(\left|x\right|=17\)
\(\Rightarrow\left[{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)
c,
|8 - 2x| + |4y - 16| = 0
\(\Rightarrow\left\{{}\begin{matrix}\left|8-2x\right|=0\\\left|4y-16\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}8-2x=0\\4y-16=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=8\\4y=16\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
d,
|x - 14| + |2y - x| = 0
\(\Rightarrow\left\{{}\begin{matrix}\left|x-14\right|=0\\\left|2y-x\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-14=0\\2y-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\2y=x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\2y=14\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\y=7\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=14\\y=7\end{matrix}\right.\)
2.Tìm x, y, z biết
a,
2.|3x| + |y + 3| + |z - y| = 0
\(\Rightarrow\left\{{}\begin{matrix}2.\left|3x\right|=0\\\left|y+3\right|=0\\\left|z-y\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x\right|=0\\y+3=0\\z-y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=0\\y=-3\\z=y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-3\\z=-3\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=0\\y=-3\\z=-3\end{matrix}\right.\)
b, (x - 3y)2 + | y + 4|= 0
\(\Rightarrow\left\{{}\begin{matrix}\left(x-3y\right)2=0\\\left|y+4\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-4\right)\\y=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
a ) x -13 = 2005
=> x = 2018
A={2018}
Vậy A có 1 phần tử
b) (x - 8)(x - 9 ) =0
\(\Rightarrow\orbr{\begin{cases}x-8=0\\x-9=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=8\\x=9\end{cases}}\)
B= {8;9}
Vậy B có 2 phần tử
a, x= 203
b, 6.x= 613+5= 618=618:6=103
c, x=1
d, x là số tự nhiên bất kì khác 0
a) \(x=2436:12=203\).
b) \(6x-5=613\)\(\Leftrightarrow6x=613+5\)\(\Leftrightarrow6x=618\)\(\Leftrightarrow x=103\).
c) \(12.\left(x-1\right)=0\)\(\Leftrightarrow x-1=0:12\)\(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=0+1=1\).
d) \(0.x=0\) suy ra x là số tự nhiên bất kì khác 0.
a, \(x^2-9=0\Rightarrow x^2=9\Rightarrow x\pm3\)
b, \(\left(x-3\right)^2-25=0\Rightarrow\left(x-3\right)^2=25\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
c, \(\left(x-3\right)\left(2x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
d, \(\left(x-3\right)x-2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
e, \(3x\left(x-1\right)-5\left(1-x\right)=0\)
\(\Rightarrow3x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
g, \(x^2+6x-7=0\)
\(\Rightarrow x^2-x+7x-7=0\)
\(\Rightarrow x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
h,\(2x^2+5x-7=0\)
\(\Rightarrow2x^2-2x+7x-7=0\)
\(\Rightarrow2x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Chúc bạn học tốt!!!
a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) vậy \(x=3;x=-3\)
b) \(\left(x-3\right)^2-25=0\Leftrightarrow\left(x-3\right)^2=25\Leftrightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
vậy \(x=8;x=-2\)
c) \(\left(x-3\right)\left(2x-5\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
vậy \(x=3;x=\dfrac{5}{2}\)
d)\(\left(x-3\right).x-2\left(x-3\right)=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) vậy \(x=2;x=3\)
e) \(3x\left(x-1\right)-5\left(1-x\right)=0\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-5}{3};x=1\)
câu e t thấy sai sai nhưng vẫn làm ; bn coi lại đề nha
g) \(x^2+6x-7=0\Leftrightarrow x^2-x+7x-7=0\)
\(\Leftrightarrow x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\) vậy \(x=-7;x=1\)
h) \(2x^2+5x-7=0\Leftrightarrow2x^2-2x+7x-7=0\)
\(\Leftrightarrow2x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(2x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+7=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-7}{2};x=1\)
1, Ta có :
a . 81 = 34 => 3x= 34 => x = 4 .
b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1
c. 23 * 2x - 1 = 64
=> 23 + ( x - 1 ) = 64 = 26
=> 3 + ( x - 1 ) = 6
=> x - 1 = 6 - 3 = 3
x = 3 + 1
x = 4
Bài 1 : \(a,\left|x-3,5\right|=7,5\)
\(\Rightarrow\orbr{\begin{cases}x-3,5=7,5\\x-3,5=-7,5\end{cases}}\Rightarrow\orbr{\begin{cases}x=11\\x=-4\end{cases}}\)
\(b,\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)
\(c,3,6-\left|x-0,4\right|=0\)
\(\Rightarrow\left|x-0,4\right|=3,6\)
\(\Rightarrow\orbr{\begin{cases}x-0,4=3,6\\x-0,4=-3,6\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3,2\end{cases}}\)
\(d,\left|x-\frac{1}{2}\right|-\frac{1}{3}=1\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{4}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{4}{3}\\x-\frac{1}{2}=-\frac{4}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{6}\\x=-\frac{5}{6}\end{cases}}\)
Bài 1: a) Do (3-2x)2 \(\ge0\) và (y-5)20 \(\ge0\)
mà (3-2x)2+(y-5)20\(\le0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(3-2x\right)^2=0\\\left(y-5\right)^{20}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3-2x=0\\y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=3-0=3\\y=0+5=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3}{2}\\y=5\end{matrix}\right.\)
Vậy: \(x=\frac{3}{2};y=5\)
c) x là các số nguyên hả bạn?
Do (x-3).(x-4)\(\le0\)
\(\Rightarrow\) Có hai trường hợp:
TH1: (x-3)(x-4)=0
Trong hai số (x-3) và (x-4) có một số bằng 0.
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0+3=3\\x=0+4=4\end{matrix}\right.\)
TH2: (x-3)(x-4)<0
Trong hai số x-3 và x-4 có một số là số nguyên dương, 1 số là số nguyên âm.
mà x-4<x-3 \(\Rightarrow\) x-4 là số nguyên âm ( x-4<0) \(\Leftrightarrow\) x<4 (1)
x-3 là số nguyên dương (x-3>0) \(\Rightarrow x>3\) (2)
Từ (1) và (2) \(\Rightarrow\) 3<x<4 mà x là các số nguyên nên x ko tm
Vậy: x\(\in\left\{3;4\right\}\)
Bài 2:
c) (x-12).(y+5)=7=1.7=7.1=-1.-7=-7.-1
\(\Rightarrow\) \(\left[{}\begin{matrix}x-12=1;y+5=7\\x-12=7;y+5=1\\x-12=-1;y+5=-7\\x-12=-7;y+5=-1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=13;y=2\\x=19;y=-4\\x=11;y=-12\\x=5;y=-6\end{matrix}\right.\)
Vậy:...
Bài 1:
\(a.\left|x\right|+\left|6\right|=\left|-27\right|\\ \Leftrightarrow\left|x\right|+6=27\\ \Leftrightarrow\left|x\right|=27-6=21\\ \Leftrightarrow\left\{{}\begin{matrix}x=-21\\x=21\end{matrix}\right.\)
a. |x||x| + |+6||+6| = |−27|
x + 6 = 27
x = 27 - 6
x = 21
Vậy x = 21
b. |−5||−5| . |x||x| = |−20|
5 . x = 20
x = 20 : 5
x 4
Vậy x = 4
c. |x| = |−17| và x > 0
|x| = 17
Vì |x| = 17
nên x = -17 hoặc 17
mà x > 0 => x = 17
Vậy x = 17 hoặc x = -17
d. |x||x| = |23||23| và x < 0
|x| = 23
Vì |x| = 23
nên x = 23 hoặc -23
mà x < 0 => x = -23
e. 12 ≤≤ |x||x| < 15
Vì 12 ≤ |x| < 15
nên x = {12; 13; 14}
Vậy x € {12; 13; 14}
f. |x| > 3
Vì |x| > 3
nên x = -2; -1; 0; 1; 2;
Vậy x € {-2; -1; 1; 2}
a. A=
{
x∈Z|−3<x≤7}
A = {-2; -1; 0; 1; 2; 3; 4; 5; 6; 7}
b. B={x∈Z|3≤|x|<7}
B = {3; 4; 5; 6}
c. C={x∈Z||x|>5}
C = {6; 7; 8; 9; ...}
Tìm số tự nhiên x, biết :
a) \(x:13=41\)
\(x=41.13\)
\(x=533\)
b) \(1428:x=14\)
\(x=1428:14\)
\(x=102\)
c) \(4x:17=0\)
\(4x=0.17\)
\(4x=0\)
\(x=0:4\)
\(x=0\)
d) \(7x-8=713\)
\(7x=713+8\)
\(7x=721\)
\(x=721:7\)
\(x=103\)
e) \(8\left(x-3\right)=0\)
\(\left(x-3\right)=0:8\)
\(x-3=0\)
\(x=0+3\)
\(x=3\)
g) \(0:x=0\)
\(x\) là số chia \(\Rightarrow x\ne0\). \(0:x=0\) \(\Rightarrow\) \(x.0=0\). Vì mọi số nhân với 0 đều bằng 0 nên \(x\in N;x\ne0\)
a) x : 13 = 41
=> x = 41 . 13
=> x = 533
b) 1428 : x = 14
=> x = 1428 : 14
=> x = 102
c) 4x : 17 = 0
=> 4x = 0 : 17
=> 4x = 0
=> x = 0 : 4
=> x = 0
d) 7x - 8 = 713
=> 7x = 713 + 8
=> 7x = 721
=> x = 721 : 7
=> x = 103
e) 8 ( x - 3 ) = 0
=> x - 3 = 0 : 8
=> x - 3 = 0
=> x = 0 + 3
=> x = 3
g) 0 : x = 0
=> x thuộc N*