Tìm x, y nguyên biết: xy+3x-y=6
Ta có: \(\left|3x-5\right|+\left|3x+1\right|=\left|5-3x\right|+\left|3x+1\right|\ge\left|5-3x+3x+1\right|=6\)
Dấu "=" xảy ra \(\Leftrightarrow\left(5-3x\right)\left(3x+1\right)\ge0\Leftrightarrow-\dfrac{1}{3}\le x\le\dfrac{5}{3}\)
Vậy \(-\dfrac{1}{3}\le x\le\dfrac{5}{3}\)
Tìm cặp số x,y nguyên biết:
\(\left|x-2y-1\right|=\dfrac{10}{\left|y-4\right|+2}\)
Với mọi x,y ta có :
\(+,\left|x-2y-1\right|\ge0\)
+, \(\left|y-4\right|+2\ge2\Leftrightarrow\dfrac{10}{\left|x-4\right|+2}\le5\)
Dấu "=" xảy ra khi :
\(\left\{{}\begin{matrix}\left|x-2y-1\right|=5\\\dfrac{10}{\left|x-4\right|+2}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=9\end{matrix}\right.\)
Vậy ..
Cho 3 số thực x, y, z thoả mãn 2x + 2y + z = 4. Tìm giá trị lớn nhất của biểu thức A = 2xy + yz + zx
Ta có : \(2x+2y+z=4\)
\(\Rightarrow z=4-2x-2y\)
Khi đó \(A=2xy+yz+zx\)
\(=2xy+\left(y+x\right)z\)
\(=2xy+\left(y+x\right)\left(4-2x-2y\right)\)
\(=2xy+4y-2xy-2y^2+4x-2x^2-2xy\)
\(=4y+4x-2y^2-2x^2-2xy\)
\(\Rightarrow2A=-4x^2-4xy+8x-4y^2+8y\)
\(=-4x^2-4x\left(y-2\right)-4y^2+8y\)
\(=-4x^2-2.2x\left(y-2\right)-\left(y-2\right)^2+\left(y-2\right)^2-4y^2+8y\)
\(=-\left(2x+y-2\right)^2-3y^2+4y+4\)
\(=-\left(2x+y-2\right)^2-3\left(y^2-\dfrac{4}{3}y-\dfrac{4}{3}\right)\)
\(=-\left(2x+y-2\right)^2-3\left(y^2-2.y.\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{4}{9}-\dfrac{4}{3}\right)\)
\(=-\left(2x+y-2\right)^2-3\left(y-\dfrac{2}{3}\right)^2+\dfrac{16}{3}\le\dfrac{16}{3}\)
\(\Rightarrow A\le\dfrac{8}{3}\)
\(Max_A=\dfrac{8}{3}\Leftrightarrow\left\{{}\begin{matrix}y-\dfrac{2}{3}=0\\2x+y-2=0\\z=4-2x-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{3}\\x=\dfrac{2}{3}\\z=\dfrac{4}{3}\end{matrix}\right.\)
Tìm x biết:
\(\left|x+2\right|+\left|x+\dfrac{3}{5}\right|=10x-\left|x+\dfrac{1}{2}\right|\)
\(pt\Leftrightarrow\left|x+2\right|+\left|x+\dfrac{3}{5}\right|+\left|x+\dfrac{1}{2}\right|=10x\)
Ta có: \(\left|x+2\right|+ \left|x+\dfrac{3}{5}\right|+\left|x+\dfrac{1}{2}\right|\ge0\Leftrightarrow10x\ge0\Leftrightarrow x\ge0\)
Khi \(x\ge0\) thì: \(x+2+x+\dfrac{3}{5}+x+\dfrac{1}{2}=10x\)
\(\Rightarrow7x+2+\dfrac{3}{5}+\dfrac{1}{2}=\dfrac{31}{10}\Leftrightarrow x=\dfrac{31}{70}\)
Tìm cặp số x,y nguyên biết:
\(\left|x-2y-1\right|+5=\dfrac{10}{\left|y-4\right|+2}\)
Tìm x,y ∈ Z biết:
\(3\left|2x+1\right|+4\left|2y-1\right|\le7\)
T đã hứa thì t sẽ làm:v
\(3\left|2x+1\right|+4\left|2y-1\right|\le7\)
\(\Rightarrow3\left|2x+1\right|\le7-4\left|2y-1\right|\le7\)
mà: \(\left\{{}\begin{matrix}3 \left|2x+1\right|\ge0\\3\left|2x+1\right|⋮3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\le3\left|2x+1\right|\le7\\3\left|2x+1\right|⋮3\end{matrix}\right.\)
Vì x nguyên nên: \(3\left|2x+1\right|\in\left\{0;3;6\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\left|2x+1\right|=0\\\left|2x+1\right|=1\\\left|2x+1\right|=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(loại\right)\\\left[{}\begin{matrix}2x+1=1\Leftrightarrow x=0\left(chọn\right)\\2x+1=-1\Leftrightarrow x=-1\left(chọn\right)\end{matrix}\right.\\\left[{}\begin{matrix}2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(loại\right)\\2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
Với \(\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) thì: \(3\left|2x+1\right|=3\Leftrightarrow4\left|2y-1\right|\le7-3=4\)
Vì \(y\in Z\) nên: \(\left[{}\begin{matrix}4\left|2y-1\right|=4\\4\left|2y-1\right|=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}2y-1=1\Leftrightarrow y=1\left(chọn\right)\\2y-1=-1\Leftrightarrow y=0\left(chọn\right)\end{matrix}\right.\\2y=1\Leftrightarrow y=\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
Vậy: \(\left(x;y\right)=\left(0;0\right);\left(0;1\right);\left(-1;1\right);\left(-1;0\right)\)
Dựa vào điều kiện: \(x;y\in Z\) là giải ra thôi bạn:v
Vì: \(x;y\in Z\) \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}0\le3\left|2x+1\right|\le7\\3\left|2x+1\right|⋮3\end{matrix}\right.\\\left\{{}\begin{matrix}0\le4\left|2y-1\right|\le7\\4\left|2y-1\right|⋮4\end{matrix}\right.\end{matrix}\right.\)
Tìm x biết:
| 2x - 6 | + 5x = 10
@saint suppapong udomkaewkanjana @Akai Haruma @ Mashiro Shiina @Nguyễn Thanh Hằng
\(\left|2x-6\right|+5x=10\)
\(\Leftrightarrow\left|2x-6\right|=10-5x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=10-5x\\2x-6=5x-10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5x=10+6\\-6+10=5x-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}7x=16\\3x=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{16}{7}\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy...
Mk cũng ở cái hoc24 này gần 2 năm rồi nhưng chưa ai nói mk xuống cấp đâu bạn à :)
\(\Leftrightarrow\left|2x-6\right|=10-5x\)
\(\left|2x-6\right|=\left[{}\begin{matrix}2x-6\left(đk:x\ge3\right)\\6-2x\left(đk:x< 3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=10-5x\left(đk:x\ge3\right)\\6-2x=10-5x\left(đk:x< 3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{7}\left(đk:x\ge3\right)\\x=\dfrac{4}{3}\left(đk:x< 3\right)\end{matrix}\right.\)
Th1 hiển nhiên hoại theo đk
x=4/3
a) tìm các số nguyên x để : B= I x-1 I + I x-2 I đạt giá trị nhỏ nhất !
b) tìm số nguyên x,y biết : xy+3x-y =6
b,xy+3x-y=6
(xy+3x)-(y+3)=3 0,5
x(y+3)-(y+3) =3
(x-1)(y+3)=3=3.1=-3.(-1) 0,5
Có 4 trường hợp xảy ra :
; ; ;
Từ đó ta tìm được 4 cặp số x; y thoả mãn là :
(x=4;y=-2) ; (x=2;y=0) ; (x=-2;y=-4) ; (x=0; y=-6) 1.0
phần a khó quá
tìm x, y nguyên biết: xy +3x-y=6
\(xy+3x-y=6\\ \Rightarrow x\left(y+3\right)-y-3=3\\ \Rightarrow x\left(y+3\right)-\left(y+3\right)=3\\ \Rightarrow\left(x-1\right)\left(y+3\right)=3\)
Ta có bảng:
x-1 | -1 | -3 | 1 | 3 |
y+3 | -3 | -1 | 3 | 1 |
x | 0 | -2 | 2 | 4 |
y | -6 | -4 | 0 | -2 |
Vậy\(\left(x,y\right)\in\left\{\left(0;-6\right);\left(-2;-4\right);\left(2;0\right);\left(4;-2\right)\right\}\)