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\(3x^2y^4\)-\(5xy^3\)-\(\dfrac{3}{2}x^2y^4\)+\(3xy^3\)+\(2xy^3\)+1=1,5\(x^2y^4\)+1>0
\(\left\{{}\begin{matrix}P\left(x\right)=x+x^2-x^3+2x^3+2=x^3+x^2+x+2\\Q\left(x\right)=1+3x-x^2-4x+x^3=x^3-x^2-x+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}P\left(x\right)+Q\left(x\right)=2x^3+3\\P\left(x\right)-Q\left(x\right)=2x^2+2x+1\end{matrix}\right.\)
a) \(2x^2-4x+7\)
\(=2\left(x^2-2x+\dfrac{7}{2}\right)\)
\(=2\left(x^2-x-x+\dfrac{7}{2}\right)\)
\(=2\left(x^2-x-x+1+\dfrac{5}{2}\right)\)
\(=2\left[\left(x-1\right)^2+\dfrac{5}{2}\right]\)
\(=2\left(x-1\right)^2+5\)
Vì \(2\left(x-1\right)^2\ge0\Rightarrow2\left(x-1\right)^2+\dfrac{5}{2}\ge\dfrac{5}{2}>0\)
\(\Rightarrow\) đt vô nghiệm.
Mấy câu kia cũng tách tương tự.
" Giữ nguyên hạng tử bậc hai chia đội hạng tử bậc nhất cân bằng hệ số để đạt được tỉ lệ thức"
Chúc bạn học tốt!!!
a)\(4x^2+4x+1=0\)
\(\Leftrightarrow4x^2+2x+2x+1=0\)
\(\Leftrightarrow2x\left(2x+1\right)+\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Leftrightarrow2x+1=0\)\(\Leftrightarrow x=-\dfrac{1}{2}\)
b)\(4x^2+5x+2=0\)
\(\Leftrightarrow4x^2+5x+\dfrac{25}{16}+\dfrac{7}{16}=0\)
\(\Leftrightarrow4\left(x^2+\dfrac{5x}{4}+\dfrac{25}{64}\right)+\dfrac{7}{16}=0\)
\(\Leftrightarrow4\left(x+\dfrac{5}{8}\right)^2+\dfrac{7}{16}>0\forall x\) ( vô nghiệm )
Sửa đề: \(C=\left(x^2y^3+x^3y^2-x^2-y^2+5\right)-\left(x^2y^3+x^3y^2+2y^2-1\right)\)
\(C=x^2y^3+x^3y^2-x^2-y^2+5-x^2y^3-x^3y^2-2y^2+1\)
\(=-3y^2-x^2+6\le6\)
Dấu '=' xảy ra khi x=y=0
cảm ơn
a/ \(\left(4x^2y^3\right)\left(x^ny^7\right)=4x^5y^{10}\)
\(\Leftrightarrow4x^{2+n}y^{3+7}=4x^5y^{10}\)
\(\Rightarrow2+n=5\Rightarrow n=3\)
Vậy \(n=3\)
b/ \(\left(-7x^4y^m\right)\left(-5x^ny^4\right)=35x^9y^{15}\)
\(\Leftrightarrow35x^{4+n}y^{m+4}=35x^9y^{15}\)
\(\Rightarrow\left[{}\begin{matrix}4+n=9\\m+4=15\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=5\\m=11\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}m=11\\n=5\end{matrix}\right.\)
a) \(\left(4x^2\times y^3\right)\left(x^n\times y^7\right)=4x^5y^{10}\)
\(\Rightarrow4\times\left(x^2\times x^n\right)\times\left(y^3\times y^7\right)=4x^5y^{10}\)
\(\Rightarrow4x^{2+x}y^{10}=4x^5y^{10}\)
\(\Rightarrow x^{2+n}=x^5\)
\(\Rightarrow2+n=5\)
\(\Rightarrow n=5-2\)
\(\Rightarrow n=3\)
Vậy \(n=3\).
b) \(\left(-7x^4y^m\right)\left(-5x^ny^4\right)=35x^9y^{15}\)
\(\Rightarrow\left[\left(-7\right)\times\left(-5\right)\right]\times\left(x^4\times x^n\right)\times\left(y^m\times y^4\right)=35x^9y^{15}\)
\(\Rightarrow35x^{4+n}y^{m+4}=35x^9y^{15}\)
\(\Rightarrow\left\{{}\begin{matrix}x^{4+n}=x^9\\y^{m+4}=y^{15}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4+n=9\\m+4=15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=9-4\\m=15-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=5\\m=9\end{matrix}\right.\)
Vậy \(m=9\) và \(n=5\).
1. a, Ta có: \(2^{24}=2^{3^8}=8^8\)
Lại có: \(3^{16}=3^{2^8}=9^8\)
Vì \(8^8< 9^8\Rightarrow2^{24}< 3^{16}\)
b, Ta có: \(5^{300}=5^{3^{100}}=125^{100}\)
Lại có: \(3^{500}=3^{5^{100}}=243^{100}\)
Vì \(125^{100}< 243^{100}\Rightarrow5^{300}< 3^{500}\)
c, Ta có: \(2^{700}=2^{7^{100}}=128^{100}\)
Lại có: \(5^{300}=5^{3^{100}}=125^{100}\)
Vì \(128^{100}>125^{100}\Rightarrow2^{700}>5^{300}\)
d, Ta có: \(2^{400}=2^{2^{200}}=4^{200}\)
\(\Rightarrow2^{400}=4^{200}\)
e, Ta có: \(99^{20}=99^{2^{10}}=9801^{10}\)
Vì \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
Bài 1:
a) Ta có: 224 = (23)8 = 88 ; 316 = (32)8 = 98
Vì 8 < 9 nên 88 < 98
Vậy 224 < 316.
b) Ta có: 5300 = (53)100 =125100 ; 3500 = (35)100 = 243100
Vì 125 < 243 nên 125100 < 243100
Vậy 5300 < 3500.
c) Ta có: 2700 = (27)100 = 128100; 5300 = (53)100 = 125100
Vì 128 > 125 nên 128100 > 125100
Vậy 2700 > 5300.
d) (làm tương tự)
Vậy 2400 = 4200.
e) (tương tự)
Vậy 9920 < 999910.
f) Ta có: 321 = 320. 3 = 910. 3 ; 231 = 230. 3 = 810. 2
Vì 910 > 810 ; 3 > 2
Nên 910. 3 > 810. 2
Vậy 321 > 231.
Bài 2: phương trình dễ ợt :v
\(A=\dfrac{4^2}{1.3}+\dfrac{4^2}{3.5}+\dfrac{4^2}{5.8}+...+\dfrac{4^2}{45.47}.\dfrac{1-3-5-...-49}{8}\)
\(A=4\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.8}+...+\dfrac{4}{45.47}\right).\dfrac{1-3-5-...-49}{8}\)\(A=4\left[2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{45}-\dfrac{1}{47}\right)\right].\dfrac{1-3-5-...-49}{8}\)\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{1-3-5-...-49}{8}\)
\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{-623}{8}\)
\(A=\dfrac{368}{47}.\dfrac{-623}{8}=\dfrac{-28658}{47}\)
a: \(B=\dfrac{3}{5}x^2+\dfrac{2}{5}x-0,5-1+\dfrac{2}{5}x-\dfrac{3}{5}x^2=-1.5\)
b: \(=1,7-12a^2-2+5a^2-7a+2.3+7a^2+7a\)
=2
c: \(=1-b^2-5b+3b^2+1+5b-2b^2=2\)