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Giải phương trình:
a) x+1 /9 + x+2 /8 = x+3 /7 + x+4 /6
b) x+43 /57 + x+46 /54 = x+49 /51 + x+52 /48
2
19 tháng 2 2017
a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+2\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
Mà \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
Vậy x = -10
b) \(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
\(\Rightarrow\left(\frac{x+43}{57}+1\right)+\left(\frac{x+46}{54}+1\right)=\left(\frac{x+49}{51}+1\right)+\left(\frac{x+52}{48}+1\right)\)
\(\Rightarrow\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
\(\Rightarrow\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
Mà \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)
\(\Rightarrow x+100=0\)
\(\Rightarrow x=-100\)
Vậy x = -100
19 tháng 2 2017
a.\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
=>\(\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
<=> \(\frac{x+1+9}{9}+\frac{x+2+8}{8}=\frac{x+3+7}{7}+\frac{x+4+6}{6}\)
<=>\(\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)
<=> \(\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
<=> \(\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
<=> x+10=0
<=> x=-10
Vậy tập nghiệm của phương trình trên là S=\(\left\{-10\right\}\)
b. \(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
=> \(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)<=>\(\frac{x+43+57}{57}+\frac{x+46+54}{54}=\frac{x+49+51}{51}+\frac{x+52+48}{48}\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
<=>(x+100)\(\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)\)=0
<=>x+100=0
<=>x= -100
Vậy tập nghiệm của phương trình trên là S=\(\left\{-100\right\}\)
31 tháng 8 2021
\(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\left(5^{128}-1\right)=2.5^{128}-2\)
31 tháng 8 2021
c: Ta có: \(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^2-1\right)\left(5^2+1\right)\cdot\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{16}-1\right)\cdot\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{64}-1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{128}-1\right)\)
\(=2\cdot5^{128}-2\)
4 tháng 11 2017
1) \(x^2+x-6=x\left(x-2\right)+3\left(x-2\right)=\left(x+3\right)\left(x-2\right)\)
2) \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3) \(x^2+2x-48=\left(x-6\right)\left(x+8\right)\)
4) \(x^2-2x-48=\left(x-8\right)\left(x+6\right)\)
5) \(x^2+x-42=\left(x-6\right)\left(x+7\right)\)
6) \(x^2-x-42=\left(x-7\right)\left(x+6\right).\)
NN
8 tháng 4 2020
a) \(4^{x+1}-4^x=48\)\(\Leftrightarrow4^x.4-4^x=48\)\(\Leftrightarrow4^x\left(4-1\right)=48\)\(\Leftrightarrow4^x.3=48\)\(\Leftrightarrow4^x=16=4^2\)\(\Leftrightarrow x=2\)
Vậy tập nghiệm của phương trình là \(S=\left\{2\right\}\)
DT
1
8 tháng 10 2017
\(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(=\left(3x-9\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(=3x^2+21x-9x-63+x^2-8x+16+48\)
\(=4x^2+4x+1\)
\(=\left(2x\right)^2+2.2x.1+1^2\)
\(=\left(2x+1\right)^2\)
\(=\left[2.\left(-\dfrac{1}{2}\right)+1\right]^2\)
= 0
CB
1
TV
0
15 tháng 3 2020
\(\frac{43-x}{57}+\frac{46-x}{54}=\frac{49-x}{51}+\frac{52-x}{48}\)
\(\Leftrightarrow\left(\frac{43-x}{57}+1\right)+\left(\frac{46-x}{54}+1\right)=\left(\frac{49-x}{51}+1\right)+\left(\frac{52-x}{48}+1\right)\)
\(\Leftrightarrow\frac{43-x+57}{57}+\frac{46-x+54}{54}=\frac{49-x+51}{51}+\frac{52-x+48}{48}\)
\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}=\frac{100-x}{51}+\frac{100-x}{48}\)
\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}-\left(\frac{100-x}{51}+\frac{100-x}{48}\right)=0\)
\(\Leftrightarrow\left(100-x\right)\left[\left(\frac{1}{57}+\frac{1}{54}\right)-\left(\frac{1}{51}+\frac{1}{48}\right)\right]=0\) (*)
Vì\(\frac{1}{57}< \frac{1}{51},\frac{1}{54}< \frac{1}{48}\Rightarrow\left(\frac{1}{57}+\frac{1}{54}\right)< \left(\frac{1}{51}+\frac{1}{48}\right)\)
\(\Rightarrow\left(\frac{1}{57}+\frac{1}{54}\right)-\left(\frac{1}{51}+\frac{1}{48}\right)< 0\)
Phương trình (*) xảy ra khi: \(100-x=0\Leftrightarrow x=100\)
Vậy phương trình có nghiệm duy nhất là x = 100
NN
0
432 là mẫu chung nha bạn
ủng hộ nha