A = -6x^2 + 3x - 1
B = -4x^2 +x
C = -49x^2 +3x-2
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ND
Nguyễn Đức Trí
VIP
18 tháng 9 2023
a) \(4x-\sqrt[]{3\left(3x-1\right)}=3x-1\)
\(\Leftrightarrow\sqrt[]{3\left(3x-1\right)}=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\3\left(3x-1\right)=\left(x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\9x-3=x^2+2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\left(a\right)\\x^2-7x+4=0\left(1\right)\end{matrix}\right.\)
Giải \(pt\left(1\right):\)
\(\Delta=49-16=33\Rightarrow\sqrt[]{\Delta}=\sqrt[]{33}\)
Phương trình (1) có 2 nghiệm phân biệt
\(\left[{}\begin{matrix}x=\dfrac{7+\sqrt[]{33}}{2}\\x=\dfrac{7-\sqrt[]{33}}{2}\end{matrix}\right.\) (thỏa \(\left(a\right)\))
29 tháng 6 2021
Bài 2 :
\(A=4x^2-2.2x.2+4+1\)
\(=\left(2x-2\right)^2+1\)
Thấy : \(\left(2x-2\right)^2\ge0\)
\(A=\left(2x-2\right)^2+1\ge1\)
Vậy \(MinA=1\Leftrightarrow x=1\)
\(B=\left(5x\right)^2-2.5x.1+1-4\)
\(=\left(5x-1\right)^2-4\)
Thấy : \(\left(5x-1\right)^2\ge0\)
\(\Rightarrow B=\left(5x-1\right)^2-4\ge-4\)
Vậy \(MinB=-4\Leftrightarrow x=\dfrac{1}{5}\)
\(C=\left(7x\right)^2-2.7x.2+4-5\)
\(=\left(7x-2\right)^2-5\)
Thấy : \(\left(7x-2\right)^2\ge0\)
\(\Rightarrow C=\left(7x-2\right)^2-5\ge-5\)
Vậy \(MinC=-5\Leftrightarrow x=\dfrac{2}{7}\)
29 tháng 6 2021
\(1.\)
\(A=-x^2-10x+1=-\left(x^2+10x-1\right)\)
\(=-\left(x^2+2.5x+5^2-5^2-1\right)=-\left[\left(x+5\right)^2-26\right]\)
\(=-\left(x+5\right)^2+26\le26\) dấu "=" xảy ra<=>x=-5
\(B=-4x^2-6x-5=-4\left(x^2+\dfrac{6}{4}x+\dfrac{5}{4}\right)\)
\(=-4\left(x^2+2.\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{11}{16}\right)\)\(=-4\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{6}\right]\le-\dfrac{11}{4}\)
\(C=-16x^2+8x-1=-16\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)
\(=-16\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)=-16\left(x-\dfrac{1}{4}\right)^2\le0\)
dấu"=" xảy ra<=>x=1/4
DA
2
19 tháng 10 2017
\(a,x^3+3x^2=4x+12\)
\(x^2\left(x+3\right)=4\left(x+3\right)\)
\(\Rightarrow\left(x+3\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm2\end{cases}}\)
\(b,49x^2=\left(3x+2\right)^2\)
\(7x=3x+2\)
\(\Rightarrow7x-3x=2\)
\(\Rightarrow4x=2\)
\(\Rightarrow x=\frac{1}{2}\)
các câu còn lại tương tự nha
19 tháng 10 2017
\(a,x^3+3x^2=4x+12\)
\(x^3+3x^2-4x-12=0\)
\(\Rightarrow x^2\left(x+3\right)-4\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\\left(x+2\right)\left(x-2\right)=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm2\end{cases}}\)
\(b,49x^2=\left(3x+2\right)^2\)
\(\Rightarrow\left(7x\right)^2=\left(3x+2\right)^2\)
\(\Rightarrow7x=3x+2\)
\(\Rightarrow7x-3x=2\)
\(\Rightarrow4x=2\)
\(\Rightarrow x=\frac{1}{2}\)
\(c,3x^2\left(x-5\right)+12\left(5-x\right)=0\)
\(3x^2\left(x-5\right)-12\left(x-5\right)=0\)
\(\left(x-5\right)\left(3x^2-12\right)=0\)
\(\Rightarrow3.\left(x-5\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\pm2\end{cases}}}\)
\(d,x^2\left(x-5\right)+45-9x=0\)
\(x^2\left(x-5\right)+9\left(5-x\right)=0\)
\(x^2\left(x-5\right)-9\left(x-5\right)=0\)
\(\left(x-5\right)\left(x^2-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2-9=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=\pm3\end{cases}}\)
AP
1
10 tháng 1 2023
a: \(=\dfrac{x-2x-1}{x+1}=\dfrac{-\left(x+1\right)}{x+1}=-1\)
b: \(=\dfrac{2+2x}{x\left(x+1\right)}=\dfrac{2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2}{x}\)
c: \(=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
tìm GTLN nha mik QuÊn