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C1 : lần trước mình giải
C2 : mình không chắc thử xem
thay x= 9 vao F ta có
F = 9^14 - 10 .9^13 + 10.9^12 - 10 .9^11 + ... +10.9^2 -10.9 + 10
= 9^14 - ( 9 + 1 ) . 9^13 + (9+1). 9^12+..+(9+1) .9^2 - (9+1)9 +10
= 9^14 - 9^14 - 9^13 + 9^13 + 9^12 -.....+9^3 + 9^2 - 9^2 - 9 + 10 = 1
Tương tự vói G , H
\(a,\frac{x-22}{8}+\frac{x-21}{9}+\frac{x-20}{10}+\frac{x-19}{11}\)
\(a,\frac{x-22}{8}+\frac{x-21}{9}+\frac{x-20}{10}+\frac{x-19}{11}=4\)
\(\Leftrightarrow\frac{x-22}{8}-1+\frac{x-21}{9}-1+\frac{x-20}{10}-1+\frac{x-19}{11}-1=0\)
\(\Leftrightarrow\frac{x-30}{8}+\frac{x-30}{9}+\frac{x-30}{10}+\frac{x-30}{11}=0\)
\(\Leftrightarrow\left(x-30\right)\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)=0\)
\(\Leftrightarrow\left(x-30\right)=0\)
\(\Leftrightarrow x=30\)
Đặt biểu thức là A, ta có:
\(A=\frac{x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)
\(\Rightarrow A.x^5=\frac{x^{45}+x^{35}+x^{25}+x^{15}+x^5}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)
\(\Rightarrow A.x^5+A=\frac{x^{45}+x^{40}+x^{35}+x^{25}+x^{15}+x^5+x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)
\(\Rightarrow A.x^5+1=1\)
\(\Rightarrow A=\frac{1}{x^5+1}\)
\(\frac{x^{24}+x^{20}+...+x^4+1}{x^{26}+x^{24}+...+x^2+1}=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^{24}+x^{20}+...+x^4+1\right)+\left(x^{26}+x^{22}+...+x^2\right)}\)
\(=1-\frac{x^2\left(x^{24}+x^{20}+...+x^4+x^1\right)}{\left(1+x^2\right)\left(x^{24}+2^{20}+...+x^4+1\right)}=1-\frac{x^2}{1+x^2}\)
\(=\frac{1+x^2-x^2}{1+x^2}=\frac{1}{1+x^2}\)
Hoặc cách khác:
\(\frac{x^{24}+x^{20}+...+x^4+1}{x^{26}+x^{24}+...+x^2+1}=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^{24}+x^{20}+...+x^4+1\right)+x^2\left(x^4+x^{20}+...+x^4+1\right)}\)
\(=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^2+1\right)\left(x^{24}+x^{20}+...+x^4+1\right)}=\frac{1}{x^2+1}\)
\(A=\frac{2\left|x-4\right|}{x^2+x-20}\)
\(x^2+x-20=x^2-4x+5x-20\)
\(=x\left(x-4\right)+5\left(x-4\right)\)
\(=\left(x-4\right)\left(x+5\right)\)
Trường hợp 1 : \(x-4\ge0\Leftrightarrow x\ge4\)
\(\Rightarrow\)\(A=\frac{2\left(x-4\right)}{\left(x-4\right)\left(x+5\right)}=\frac{2}{x+5}\)
Trường hợp 2 : \(x-4< 0\Leftrightarrow x< 4\)
\(\Rightarrow\)\(A=\frac{2\left(4-x\right)}{\left(x-4\right)\left(x+5\right)}=\frac{-2}{x+5}\)
A = 2|x-4|/(x-4).(x+5)
Nếu x<4 thì |x-4| = -(x+4) => A = -2/x+%
Nếu x>=4 thì |x-4|=x-4 => A = 2/x+5
Vậy ........
k mk nha
noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
1) Ta có: \(5\left(x-2\right)=3x+10\)
\(\Leftrightarrow5x-10-3x-10=0\)
\(\Leftrightarrow2x-20=0\)
\(\Leftrightarrow2\left(x-10\right)=0\)
Vì 2>0
nên x-10=0
hay x=10
Vậy: x=10
2) Ta có: \(x^2\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x^2\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)
Vậy: x∈{-2;2;5}
3) Ta có: \(\frac{3x+1}{4}+\frac{8x-21}{20}=\frac{3\left(x+2\right)}{5}-2\)
\(\Leftrightarrow\frac{5\left(3x+1\right)}{20}+\frac{8x-21}{20}-\frac{12\left(x+2\right)}{20}+\frac{40}{20}=0\)
\(\Leftrightarrow15x+5+8x-21-12\left(x+2\right)+40=0\)
\(\Leftrightarrow15x+5-8x-21-12x-24+40=0\)
\(\Leftrightarrow-5x=0\)
hay x=0
Vậy: x=0
4) ĐKXĐ: x≠5; x≠-5
Ta có: \(\frac{3}{4x-20}+\frac{7}{6x+30}=\frac{15}{2x^2-50}\)
\(\Leftrightarrow\frac{3}{4\left(x-5\right)}+\frac{7}{6\left(x+5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}+\frac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}-\frac{180}{12\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow9x+45+14x-70-180=0\)
\(\Leftrightarrow23x-205=0\)
\(\Leftrightarrow23x=205\)
hay \(x=\frac{205}{23}\)(tm)
Vậy: \(x=\frac{205}{23}\)