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\(a,2004^2-16=2004^2-4^2=\left(2004-4\right)\left(2004+4\right)=2000.2008=4016000\)\(b,892^2+892.216+108^2=892^2+2.892.108+108^2=\left(892+108\right)^2=1000^2=1000000\)\(c,10,2.9,8-9,8.02+10,2^2-10,2.0,2=9,8\left(10,2-0,2\right)+10,2\left(10,2-0,2\right)=9,8.10+10,2.10=98+102=200\)\(d,36^2+26^2-52.36=36^2-2.36.26+26^2=\left(36-26\right)^2=10^2=100\)\(e,99^3+1+3.\left(99^2+99\right)=99^3+3.99^2+3.99+1^3=\left(99+1\right)^3=100^3=1000000\)
\(f.37.43=\left(40-3\right)\left(40+3\right)=40^2-3^2=1600-9=1591\)
a: \(2004^2-16=2000\cdot2008=4016000\)
b: \(892^2+892\cdot216+108^2=1000^2=1000000\)
c: \(36^2-52\cdot36+26^2=10^2=100\)
a.
\(\dfrac{5x-17}{14}+\dfrac{x-3}{26}>\dfrac{29-9x}{91}\)
\(\Leftrightarrow13\left(5x-17\right)+7\left(x-3\right)>2\left(29-9x\right)\)
\(\Leftrightarrow65x-221+7x-21>58-18x\)
\(\Leftrightarrow65x+7x+18x>58+21+221\)
\(\Leftrightarrow90x>300\)
\(\Leftrightarrow x>\dfrac{10}{3}\)
b)
\(\dfrac{8x-1}{9}+\dfrac{3x-2}{4}< \dfrac{43+8x}{12}+\dfrac{35x}{36}\)
\(\Leftrightarrow4\left(8x-1\right)+9\left(3x-2\right)< 3\left(43+8x\right)+35x\)
\(\Leftrightarrow32x-4+27x-18< 129+24x+35x\)
\(\Leftrightarrow32x+27x-24x-35x< 129+18+4\)
\(\Leftrightarrow0x< 151\) ( luôn đúng)
Vậy bất pt vô số nghiệm
\(VT=\frac{\left(43+17\right)\left(43^2-43.17+17^2\right)}{\left(43+26\right)\left(43^2-43.26+26^2\right)}=\frac{\left(43+17\right)\left[43^2-17\left(43-17\right)\right]}{\left(43+26\right)\left[43^2-26\left(43-26\right)\right]}.\)
\(VT=\frac{\left(43+17\right)\left(43^2-17.26\right)}{\left(43+26\right)\left(43^2-26.17\right)}=\frac{43+17}{43+26}=VP\left(dpcm\right)\)
\(6^2.6^4-4^3\left(3^6-1\right)\)
\(=6^6-\left(2^2\right)^3\left(3^6-1\right)\)
\(=\left(2.3\right)^6-2^6\left(3^6-1\right)\)
\(=2^6.3^6-2^6.3^6+2^6\)
\(=2^6\)
\(=64\)
\(6^2.6^4-4^3\left(3^6-1\right)\)
\(=6^{2+4}-\left(2^2\right)^3\left(3^6-1\right)\)
\(=6^6-\left(2^6.3^6-2^6\right)\)
\(=6^6-\left(6^6-2^6\right)=6^6-6^6+2^6=2^6=64\)