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thay 11=x+1 ta có:
f(x)= \(x^{10}\)-11\(x^9\)+11\(x^8\)-11\(x^7\)+....+11\(x^2\)-11x+100
=\(x^{10}\)-(x+1)\(x^9\)+(x+1)\(x^8\)-(x+1)\(x^7\)+...+(x+1)\(x^2\)-(x+1)x+100
=\(x^{10}\)-\(x^{10}\)-\(x^9\)+\(x^9\)+\(x^8\)-\(x^8\)-\(x^7\)+......+\(x^3\)+\(x^2\)-\(x^2\)-x+100
=-x+100
=> f(10)=-10+100=90
Thay 11 = x + 1 ta có:
f(x) = \(x^{10}-11x^9+11x^8-11x^7+...+11x^2-11x+100\)
\(=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x^2-\left(x+1\right)x+100\)
= \(x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2-x+100\)
= -x+100
=>f(10)= - 10 + 100 = 90
\(\frac{11x-2}{7x-5}=\frac{11}{8}\)
\(\Leftrightarrow8\left(11x-2\right)=11\left(7x-5\right)\)
\(\Leftrightarrow88x-16=77x-55\)
\(\Leftrightarrow88x-77x=-55+16\)
\(\Leftrightarrow11x=-39\)
\(\Leftrightarrow x=\frac{-39}{11}\)
\(\frac{12x-2}{7x-8}=\frac{11}{8}\)
\(\Rightarrow8\left(12x-2\right)=11\left(7x-8\right)\)
\(\Rightarrow96x-16=77x-88\)
\(\Rightarrow96x-77x=-88+16\)
\(\Rightarrow19x=-72\)
\(\Rightarrow x=-\frac{72}{19}\)
\(\left(11x-5\right)-\left(2x-7\right)=5x-\left(8-7x\right)\Leftrightarrow11x-5-2x+7=5x-8+7x\)
\(\Leftrightarrow9x+2=12x-8\Leftrightarrow-3x+10=0\)
\(\Leftrightarrow-3x=-10\Leftrightarrow x=\frac{10}{3}\)
\(\left(11x-5\right)-\left(2x-7\right)=5x-\left(8-7x\right)\)
\(\Leftrightarrow11x-5-2x+7=5x-8+7x\)
\(\Leftrightarrow11x-2x-5x-7x=-8+5-7\)
\(\Leftrightarrow-3x=-10\)
\(\Leftrightarrow x=\frac{10}{3}\)
đề sai
nếu mà duy trì đề đó thì TA có:
11x - 2 = 11
11x = 11 + 2 = 13
x = 13 : 11
x = 1,181818182
_____________
7x + 5 = 8
7x = 8 - 5 = 3
x = 3 : 7
x = 0,428571428
mk nghĩ là nhân tích chéo.
(11x-2).8=11.(7x+5)
8.11x-2.8=11.7x+11.5
8.11x-11.7x=55+16
(8.11)-(11x-7x)=71
88-4x=71 4x=88-71 4x=17 x=17:4 x=17/4