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NV
1
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 5 2020
Đặt \(\left\{{}\begin{matrix}x-9=a\\x-10=b\end{matrix}\right.\)
\(a^4+b^4=\left(a+b\right)^4\)
\(\Leftrightarrow a^4+b^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
\(\Leftrightarrow2ab\left(2a^2+3ab+2b^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)
8 tháng 4 2020
a) \(\left(x+2\right)^2-9\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left(3x-6\right)^2=0\)
\(\Leftrightarrow\left(x+2+3x-6\right)\left(x+2-3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x-4\right)=0\\\left(8-2x\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
b)\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\frac{23}{7}\end{matrix}\right.\)
c) \(\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\)
\(\Leftrightarrow\left(5x^2-2x+10-3x^2-10x+8\right)\left(5x^2-2x+10+3x^2+10x-8\right)=0\)
\(\Leftrightarrow\left(2x^2-5x+18\right)\left(8x^2+8x+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=3\end{matrix}\right.\)
27 tháng 11 2018
a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\)
\(\Leftrightarrow4x=7\Leftrightarrow x=1,75\)
27 tháng 11 2018
\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10.\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\)
\(\Leftrightarrow24x=27\Leftrightarrow x=1,125\)
31 tháng 5 2018
3) \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
S=\(\left\{6;1\right\}\)
\(\)
KB
26 tháng 8 2018
\(A=19-3x^2-24x\)
\(\Leftrightarrow A=-3\left(x^2+8x-\dfrac{19}{3}\right)\)
\(\Leftrightarrow A=-3\left(x^2+2x.4+16-\dfrac{67}{3}\right)\)
\(\Leftrightarrow A=-3\left[\left(x+4\right)^2-\dfrac{67}{3}\right]\)
\(\Leftrightarrow A=-3\left(x+4\right)^2+67\le67\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow-3\left(x+4\right)^2=0\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
Vậy Max A là : 67 \(\Leftrightarrow x=-4\)
\(B=-x^2+6x-23\)
\(\Leftrightarrow B=-\left(x^2-6x+9\right)-14\)
\(\Leftrightarrow B=-\left(x-3\right)^2-14\le-14\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy Max B là : \(-14\Leftrightarrow x=3\)
KB
26 tháng 8 2018
\(C=4\left(x-1\right)^2-9\left(x+2\right)^2\)
\(\Leftrightarrow C=4x^2-8x+4-9x^2-36x-36\)
\(\Leftrightarrow C=-5x^2-44x-32\)
\(\Leftrightarrow C=-5\left(x^2+\dfrac{44}{5}x+\dfrac{32}{5}\right)\)
\(\Leftrightarrow C=-5\left(x^2+2x.\dfrac{22}{5}+\dfrac{484}{25}\right)+64,8\)
\(\Leftrightarrow C=-5\left(x+\dfrac{22}{5}\right)^2+64,8\le64,8\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow-5\left(x+\dfrac{22}{5}\right)^2=0\Leftrightarrow\left(x+\dfrac{22}{5}\right)^2=0\Leftrightarrow x+\dfrac{22}{5}=0\)
\(\Leftrightarrow x=-\dfrac{22}{5}\)
Vậy Max C là : 64 , 8 \(\Leftrightarrow x=-\dfrac{22}{5}\)
\(E=\left(x+2\right)^2-2x^2+8\)
\(\Leftrightarrow E=x^2+4x+4-2x^2+8\)
\(\Leftrightarrow E=-x^2+4x+12\)
\(\Leftrightarrow E=-\left(x^2-4x+4\right)+16\)
\(\Leftrightarrow E=-\left(x-2\right)^2+16\le16\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow-\left(x-2\right)^2=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy Max E là : \(16\Leftrightarrow x=2\)
19 tháng 2 2019
1) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)
2) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)
3) \(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)
27 tháng 11 2018
a) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\Leftrightarrow x=\dfrac{7}{4}\)
b) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\Leftrightarrow x=\dfrac{9}{8}\)
c) đề có sai ko vậy bạn :
\(\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)
\(\Leftrightarrow\left(x^2-8x+16\right)-\left(x^2-4\right)=6\)
\(\Leftrightarrow x^2-8x+16-x^2+4-6=0\)
\(\Leftrightarrow-8x+14=0\Leftrightarrow x=\dfrac{7}{4}\)
d) \(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(\Leftrightarrow9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)
\(\Leftrightarrow9x^2+18x+9-9x^2+4-10=0\)
\(\Leftrightarrow18x+3=0\Leftrightarrow x=\dfrac{1}{6}\)
