1.C/M: a2+5b2-(3a+b)\(\ge\)3ab-5
2. Tìm các nghiệm nguyên của pt: 2x2+3y2+4x=19
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T
4 tháng 9 2023
\(a,a^2-2a-4b^2-4b\)
\(=\left(a^2-4b^2\right)-\left(2a+4b\right)\)
\(=\left(a-2b\right)\left(a+2b\right)-2\left(a+2b\right)\)
\(=\left(a+2b\right)\left(a-2b-2\right)\)
\(b,x^3-2x^2+4x-8\)
\(=x^2\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4\right)\)
\(c,x^3+36x-12x^2\)
\(=x^3-6x^2-6x^2+36x\)
\(=x^2\left(x-6\right)-6x\left(x-6\right)\)
\(=\left(x-6\right)\left(x^2-6x\right)\)
\(=x\left(x-6\right)^2\)
\(d,5a^2+3\left(a+b\right)^2-5b^2\)
\(=\left(5a^2-5b^2\right)+3\left(a+b\right)^2\)
\(=5\left(a^2-b^2\right)+3\left(a+b\right)^2\)
\(=5\left(a-b\right)\left(a+b\right)+3\left(a+b\right)^2\)
\(=\left(a+b\right)\left[5\left(a-b\right)+3\left(a+b\right)\right]\)
\(=\left(a+b\right)\left(5a-5b+3a+3b\right)\)
\(=\left(a+b\right)\left(8a-2b\right)\)
\(=2\left(a+b\right)\left(4a-b\right)\)
\(e,x^3-3x^2+3x-1-y^3\)
\(=\left(x^3-3x^2+3x-1\right)-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)
\(=\left(x-y-1\right)\left(x^2-2x+1+xy-y+y^2\right)\)
\(=\left(x-y-1\right)\left(x^2+y^2-xy-y+1\right)\)
#Urushi☕
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
4 tháng 9 2023
\(c.\\ x^3+36x-12x^2\\ =x\left(x^2-12x+36\right)\\ =x.\left(x^2-2.x.6+6^2\right)\\ =x.\left(x-6\right)^2\\ ---\\ d.\\ 5a^2+3\left(a+b\right)^2-5b^2\\ =\left(5a^2-5b^2\right)+3\left(a+b\right)^2\\ =5.\left(a^2-b^2\right)+3.\left(a+b\right)\left(a+b\right)\\ =5\left(a+b\right)\left(a-b\right)+3\left(a+b\right)\left(a+b\right)\\ =\left(a+b\right)\left(5a-5b+3a+3b\right)\\ =\left(a+b\right)\left(8a-2b\right)\\ =2\left(a+b\right)\left(4a-b\right)\)
\(e.\\ x^3-3x^2+3x-1-y^3\\ =\left(x-1\right)^3-y^3\\ =\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right).y+y^2\right]\\ =\left(x-y-1\right).\left[\left(x^2-2x+1\right)+y\left(x+y-1\right)\right]\)
13 tháng 3 2023
a: Thay x=5 vào pt, ta được:
5^2-2(m-1)*5+m^2-4m+3=0
=>m^2-4m+3+25-10m+10=0
=>m^2-14m+38=0
=>(m-7)^2=11
=>\(m=\pm\sqrt{11}+7\)
b: x1+x2=2m-2
x1*x2=m^2-4m+3
(x1+x2)^2-4x1x2
=4m^2-8m+4-4m^2+4m-6
=-4m-2
(x1+x2)^2-4x1x2+2(x1+x2)
=-4m-2+4m-4=-6
5 tháng 10 2021
\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)
\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)
5 tháng 10 2021
\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)
Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)
HY
4
21 tháng 5 2015
\(\Leftrightarrow4x^2+8x+4=42-6y^2\)
\(\Rightarrow\left(2x+2\right)^2=6\left(7-y^2\right)\)
Vì \(\left(2x+2\right)^2\ge0\) \(\Rightarrow7-y^2\ge0\)\(\Rightarrow y^2\le7\)
Mà \(y\in Z\) \(\Rightarrow y=0\); +-1 ; +-2 \(\Rightarrow\) các gt tương ứng của x
đúng nha
bài này cũng dễ
11 tháng 11 2021
\(1,\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(-3\right)^2-4\left(-2\right)\left(-m+1\right)>0\\x_1+x_2=\dfrac{3}{-2}< 0\\x_1x_2=\dfrac{-m+1}{-2}>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}17-8m>0\\-m+1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{17}{8}\\m>1\end{matrix}\right.\Leftrightarrow1< m< \dfrac{17}{8}\)
\(2,\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(-4\right)^2-4\left(-3\right)\left(-2m+1\right)\ge0\\x_1+x_2=\dfrac{4}{-3}< 0\\x_1x_2=\dfrac{-2m+1}{-3}>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28-24m\ge0\\-2m+1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le\dfrac{7}{6}\\m>\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< m\le\dfrac{7}{6}\)
8 tháng 5 2022
\(M=\dfrac{1}{\dfrac{c}{a}+\dfrac{2a}{b}+3}+\dfrac{1}{\dfrac{a}{b}+\dfrac{2b}{c}+3}+\dfrac{1}{\dfrac{b}{c}+\dfrac{2c}{a}+3}\)
\(đặt\left(\dfrac{a}{b};\dfrac{b}{c};\dfrac{c}{a}\right)=\left(x;y;z\right)\Rightarrow xyz=1\left(x;y;z>0\right)\)
\(M=\dfrac{1}{z+2x+3}+\dfrac{1}{x+2y+3}+\dfrac{1}{y+2z+3}\)
\(ta\) \(đi\) \(cminh:A\le\dfrac{1}{2}\)
có:
\(\dfrac{1}{z+2x+3}\le\dfrac{1}{6}\Leftrightarrow z+2x+3\ge6\Leftrightarrow2x+z\ge3\)
\(\dfrac{1}{x+2y+3}\le\dfrac{1}{6}\Leftrightarrow x+2y\ge3\)
\(\dfrac{1}{y+2z+3}\le\dfrac{1}{6}\Rightarrow y+2z\ge3\)
\(cộng\) \(vế\Rightarrow2x+z+2y+x+2z+y\ge9\Leftrightarrow x+y+z\ge3\left(đúng\right)\)
\(do:x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow A\le\dfrac{1}{2}dấu"="\Leftrightarrow x=y=z=1\Rightarrow a=b=c\)
1. Ta có:
\(a^2+5b^2-\left(3a+b\right)\ge3ab-5\)
\(\Leftrightarrow2a^2+10b^2-6a-2b-6ab+10\ge0\)
\(\Leftrightarrow a^2-6ab+9b^2+a^2-6a+9+b^2-2b+1\ge0\)
\(\Leftrightarrow\left(a-3b\right)^2+\left(a-3\right)^2+\left(b-1\right)^2\ge0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}a=3\\b=1\end{cases}}\)
2. Giải:
Ta có: \(2x^2+3y^2+4x=19\)
\(\Leftrightarrow2x^2+4x+2=21-3y^2\)
\(\Leftrightarrow2\left(x+1\right)^2=3\left(7-y^2\right)\left(1\right)\)
Xét thấy \(VT⋮2\Leftrightarrow3\left(7-y^2\right)⋮2\Leftrightarrow y\) lẻ (2)
Mặt khác \(VT\ge0\Leftrightarrow3\left(7-y^2\right)\ge0\Leftrightarrow y^2\le7\) (3)
Kết hợp (2) và (3) suy ra:
\(y^2=1\) Thay vào \(\left(1\right)\) ta có:
\(2\left(x+1\right)^2=18\). Vậy ta tính được các nghiệm:
\(\left(x,y\right)=\left(2;1\right);\left(2;-1\right);\left(-4;-1\right);\left(-4;1\right)\)