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Giả sử như x/2=y/5=-z/2
Đặt x/2=y/5=z/-2=k
=>x=2k; y=5k; z=-2k
\(\left(2x+5y-2z\right)^2=\left(4k+25k+4k\right)^2=\left(41k\right)^2\)
\(33\left(x^2+y^2+z^2\right)=33\left(4k^2+25k^2-4k^2\right)\)
\(=33\cdot25k^2< >\left(41k\right)^2\)
=>Đề sai rồi bạn
1
a,
=(202+54).(202-54)+256.352
=37888+256.352
=37888+90112
=128000
b,
=621-769.373-21904
=621-286837-21904
=-308120
c,
42^2-10^2/(36,5)^2-(27,5)^2
=(42-10).(42+10)/(36,5-27,5).(27,5+36,5)
=1664/576=2(8)
1. Tính nhanh :
a) \(202^2-54^2+256.352\)
\(=\left(202-54\right).\left(202+54\right)+256.352\)
\(=148.256+256.352\)
\(=256.\left(148+252\right)=256.400=102400\)
\(e,\)
\(\left(\dfrac{1}{3}a^3b+\dfrac{1}{3}a^2b^2-\dfrac{1}{4}ab^3\right):5ab\)
\(=\dfrac{1}{15}a^2+\dfrac{1}{15}ab-\dfrac{1}{20}b^2\)
\(f,\)
\(\left(-\dfrac{2}{3}x^5y^2+\dfrac{3}{4}x^4y^3-\dfrac{4}{5}x^3y^4\right):6x^2y^2\)
\(=-\dfrac{1}{9}x^3+\dfrac{1}{8}x^2y-\dfrac{2}{15}xy^2\)
\(g,\)
\(\left(\dfrac{3}{4}a^6b^3+\dfrac{6}{5}a^3b^4-\dfrac{5}{10}ab^5\right):\left(\dfrac{3}{5}ab^3\right)\)
\(=\dfrac{5}{4}a^5+2a^2b-\dfrac{5}{6}b^2\)
a) \(\left(\frac{1}{x}+2\right)=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)\left(x^2+1\right)-\left(\frac{1}{x}+2\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)x^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}+2=0\\x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=0\left(L\right)\end{cases}}\)
Vậy \(x=-\frac{1}{2}\)
s e thấy == câu này mọi ngừi ko tl vậy :v ( bài này cs cần đk ko -.- e chưa hc nên ko nắm chắc , kệ đi , cứ lm )
\(a,\left(\frac{1}{x}+2\right)=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(1+2x=x\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(1+2x=x^2+1+2x^3+2x\)
\(2x=x^2+2x^3+2x\)
\(0=x^2+2x^3\)
\(0=x^2\left(1+2x\right)\)
\(x=0;-\frac{1}{2}\)
a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
\(\left(36,5^2-27,5^2\right):\left(\dfrac{57^3+33^3}{90}-57.33\right)\)
\(=\left(35,5-27,5\right)\left(36,5+27,5\right):\left(\dfrac{\left(57+33\right)\left(57^2-57.33+33^2\right)}{90}-57.33\right)\)
\(=9.64:\left(\dfrac{90.\left(57^2-57.33+33^2\right)}{90}-57.33\right)\)
\(=576:\left(57^2-57.33+33^2-57.33\right)\)
\(=576:\left(57^2-2.57.33+33^2\right)\)
\(=576:\left(57-33\right)^2\)
\(=576:24^2\)
\(=576:576\)
\(=1\)