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\(4^{x+2}.5.4=1280\Leftrightarrow4^{x+2}=\frac{1280}{5.4}\Leftrightarrow4^{x+2}=64\Leftrightarrow4^{x+2}=4^3\)
\(\Rightarrow x+2=3\Leftrightarrow x=3-2\Leftrightarrow x=1\)
4x+2.5.4=1280
4x+2.20=1280
4x+2 =1280:20
4x.42 =64
4x.16=64
4x =64:16
4x=4
=>x=1
\(2^x+2^{x+1}=96\)
\(\Rightarrow2^x\left(1+2\right)=96\)
\(\Rightarrow2^x.3=96\)
\(2^x=96:3\)
\(2^x=32\)
\(\Rightarrow2^x=2^5\)
\(x=5\)
Nhiều câu quá >.<
a/ \(2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20.\)
\(2x^2+10x=x^2+6x+9+x^2-2x+1+20.\)
\(10x=4x+30\)
\(6x=30\Rightarrow x=5\)
các câu còn lại tương tự
\(a,2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20\)
\(\Leftrightarrow2x^2+10x=x^2+6x+9+x^2-2x+1+20\)
\(\Leftrightarrow2x^2+10x=2x^2+4x+30\)
\(\Leftrightarrow2x^2+10x-2x^2-4x=30\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)
Vậy ...........
\(b,\left(2x-2\right)^2=\left(x+1\right)^2+3\left(x-2\right)\left(x+5\right)\)
\(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3x^2+15x-6x-30\)
\(\Leftrightarrow4x^2-8x+4=4x^2+11x-29\)
\(\Leftrightarrow4x^2-8x-4x^2-11x=-29-4\)
\(\Leftrightarrow-19x=-33\)
\(\Leftrightarrow x=\frac{33}{19}\)
Vậy...........
\(c,\left(x-1\right)^2+\left(x+3\right)^2=2\left(x-2\right)\left(x+1\right)+38\)
\(\Leftrightarrow x^2-2x+1+x^2+6x+9=2x^2+2x-4x-4+38\)
\(\Leftrightarrow2x^2+4x+10=2x^2-2x+34\)
\(\Leftrightarrow2x^2+4x-2x^2+2x=34-10\)
\(\Leftrightarrow6x=24\)
\(\Leftrightarrow x=4\)
Vậy.............
\(d,\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-18\)
\(\Leftrightarrow x^3+6x+12x+8-\left(x^3-6x+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x+12x+8-x^3+6x-12x+8=12x^2-12x-8\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy............
làm được câu a thui
\(\frac{1}{100}=\frac{1}{10.10}\);\(\frac{1}{90}=\frac{1}{9.10}\);...
Suy ra \(\frac{1}{10.10}-\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-...-\frac{1}{1.2}\)
\(\frac{1}{10}-\frac{1}{10}-\frac{1}{10}-\frac{1}{9}-...-1-\frac{1}{2}\)
\(\frac{1}{10}-\frac{1}{2}\)
\(-\frac{4}{10}\)
2.
\(\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left(65\cdot111-13\cdot15\cdot37\right)\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left(65\cdot111-13\cdot5\cdot3\cdot37\right)\\=\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left[65\cdot111-\left(13\cdot5\right)\cdot\left(3\cdot37\right)\right]\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left[65\cdot111-65\cdot111\right]\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot0\\ =0\)
2.(2x+3+1)=1280
2.(2x+4)=1280
4x+8=1280
4x =1280-8
4x =1272
x =318