tìm X thuộc Z
d,(x-2).(2y+1)=17
e,22x-1:4=83
K
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CM
2 tháng 2 2018
ĐK: x ≥ −3; y ≥ −1
Ta có:
x + 3 − 2 y + 1 = 2 2 x + 3 + y + 1 = 4 ⇔ 2 x + 3 − 4 y + 1 = 4 2 x + 3 + y + 1 = 4 ⇔ x + 3 − 2 y + 1 = 2 − 5 y + 1 = 0 ⇔ y = − 1 x + 3 − 2. − 1 + 1 = 2 ⇔ y = − 1 x + 3 = 2 ⇔ y = − 1 x + 3 = 4 ⇔ y = − 1 x = 1 t m
Vậy hệ phương trình có nghiệm duy nhất (x; y) = (1; −1)
Nên x + y = 1 + (−1) = 0
Đáp án: B
14 tháng 10 2021
\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}=0\left(x\ne1\right)\\ \Leftrightarrow x=0\)
\(d,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\left(\dfrac{2}{\sqrt{x}+1}>0\right)\\ e,P=1-\dfrac{2}{\sqrt{x}+1}\\ \sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-\dfrac{2}{1}=-2\\ \Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-\left(-2\right)=3\)
Dấu \("="\Leftrightarrow x=0\)
14 tháng 10 2021
a) ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)
\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)
c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Kết hợp đk:
\(\Leftrightarrow x\in\left\{0\right\}\)
d) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)
e) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Do \(\sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-2\)
\(\Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)
\(minP=-1\Leftrightarrow x=0\)
24 tháng 6 2021
`a)C=((2x^2+1)/(x^3-1)-1/(x-1)):(1-(x^2-2)/(x^2+x+1))`
`ĐK:x ne 1`
`C=((2x^2+1-x^2-x-1)/(x^3-1)):((x^2+x+1-x^2+2)/(x^2+x+1))`
`C=((x^2-x)/(x^3-1)):((x+3)/(x^2+x+1))`
`C=x/(x^2+x+1)*(x^2+x+1)/(x+3)`
`C=x/(x+3)`
`b)|1-x|+2=3(x+1)`
`<=>|1-x|+2=3x+3`
`<=>|1-x|=3x+1(x>=-1/3)`
`**1-x=3x+1`
`<=>4x=0<=>x=0(tmđk)`
`**x-1=3x+1`
`<=>2x=-2`
`<=>x=-1(l)`
Thay `x=0` vào C
`=>C=0`
`c)C in ZZ`
`=>x vdots x+3`
`=>x+3-3 vdots x+3`
`=>3 vdots x+3`
`=>x+3 in Ư(3)={+-1,+-3}`
`=>x in {-2,-4,0,-6}`
`d)|C|>C`
Mà `|C|>=0`
`=>C<0`
`<=>x/(x+3)<0`
Để 1 p/s `<=0` thì tử và mẫu trái dấu mà `x<x+3`
`=>` \(\begin{cases}x<0\\x+3>0\\\end{cases}\)
`<=>` \(\begin{cases}x>-3\\x<0\\\end{cases}\)
`<=>-3<x<0`
NV
0
11 tháng 12 2023
1: \(y=x+\dfrac{4}{\left(x-2\right)^2}\)
\(\Leftrightarrow y'=1+\left(\dfrac{4}{\left(x-2\right)^2}\right)'\)
=>\(y'=1+\dfrac{4'\left(x-2\right)^2-4\left[\left(x-2\right)^2\right]'}{\left(x-2\right)^4}\)
=>\(y'=1+\dfrac{-4\cdot2\cdot\left(x-2\right)'\left(x-2\right)}{\left(x-2\right)^4}\)
=>\(y'=1-\dfrac{8}{\left(x-2\right)^3}\)
Đặt y'=0
=>\(\dfrac{8}{\left(x-2\right)^3}=1\)
=>\(\left(x-2\right)^3=8\)
=>x-2=2
=>x=4
Đặt \(f\left(x\right)=x+\dfrac{4}{\left(x-2\right)^2}\)
\(f\left(4\right)=4+\dfrac{4}{\left(4-2\right)^2}=4+1=5\)
\(f\left(0\right)=0+\dfrac{4}{\left(0-2\right)^2}=0+\dfrac{4}{4}=1\)
\(f\left(5\right)=5+\dfrac{4}{\left(5-2\right)^2}=5+\dfrac{4}{9}=\dfrac{49}{9}\)
Vì f(0)<f(4)<f(5)
nên \(f\left(x\right)_{max\left[0;5\right]\backslash\left\{2\right\}}=f\left(5\right)=\dfrac{49}{9}\) và \(f\left(x\right)_{min\left[0;5\right]\backslash\left\{2\right\}}=1\)
2: \(y=cos^22x-sinx\cdot cosx+4\)
\(=1-sin^22x-\dfrac{1}{2}\cdot sin2x+4\)
\(=-sin^22x-\dfrac{1}{2}\cdot sin2x+5\)
\(=-\left(sin^22x+\dfrac{1}{2}\cdot sin2x-5\right)\)
\(=-\left(sin^22x+2\cdot sin2x\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{81}{16}\right)\)
\(=-\left(sin2x+\dfrac{1}{4}\right)^2+\dfrac{81}{16}\)
\(-1< =sin2x< =1\)
=>\(-\dfrac{3}{4}< =sin2x+\dfrac{1}{4}< =\dfrac{5}{4}\)
=>\(0< =\left(sin2x+\dfrac{1}{4}\right)^2< =\dfrac{25}{16}\)
=>\(0>=-\left(sin2x+\dfrac{1}{4}\right)^2>=-\dfrac{25}{16}\)
=>\(\dfrac{81}{16}>=-sin\left(2x+\dfrac{1}{4}\right)^2+\dfrac{81}{16}>=-\dfrac{25}{16}+\dfrac{81}{16}=\dfrac{7}{2}\)
=>\(\dfrac{81}{16}>=y>=\dfrac{7}{2}\)
\(y_{min}=\dfrac{7}{2}\) khi \(sin2x+\dfrac{1}{4}=\dfrac{5}{4}\)
=>\(sin2x=1\)
=>\(2x=\dfrac{\Omega}{2}+k2\Omega\)
=>\(x=\dfrac{\Omega}{4}+k\Omega\)
\(y_{max}=\dfrac{81}{16}\) khi sin 2x=-1
=>\(2x=-\dfrac{\Omega}{2}+k2\Omega\)
=>\(x=-\dfrac{\Omega}{4}+k\Omega\)
NN
0
(x - 2)(2y + 1) = 17
=> x - 2 và 2y + 1 thuộc Ư(17) = {-1; 1; -17; 17}
ta có bảng :
vậy ta có các cặp số x; y thỏa mãn : (1; -9); (3; 8); (-15; -1); (19; 0)
22x - 1 : 4 = 83
=> 22x - 1 = (23)3.4
=> 22x - 1 = 29.22
=> 22x - 1 = 211
=> 2x - 1 = 11
=> 2x = 12
=> x = 6
vậy_
Có( x-2).(2y+1)=17
Do x, y €Z => x-2 và 2y+1 €Z
=>x-2 và 2y+1 € Ư (17)
<=> x-2 và 2y +1€{-17,-1,1,17}
Ta có bảng sau:
Vậy các cấp gtrị (x, y) t/m đề bài là (-15,-1);(1,-9);(3,8);(19,0).