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Lí do = 0 nè:
\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-5}{39}+\frac{x-69}{49}\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}=0+\left(\frac{x-5}{39}+\frac{x-69}{49}\right)\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}-\left(\frac{x-5}{39}+\frac{x-69}{49}\right)=0\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}-\frac{x-5}{39}-\frac{x-69}{49}=0\)
a,x/8=2/x
=>x2=16
=>x=-4;4
vậy x=-4;4
b,\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-59}{39}+\frac{x-69}{49}\)
\(\Rightarrow\left(\frac{x-39}{19}+1\right)+\left(\frac{x-49}{29}+1\right)=\left(\frac{x-59}{39}+1\right)+\left(\frac{x-69}{49}+1\right)\)
\(\Rightarrow\frac{x-20}{19}+\frac{x-20}{29}=\frac{x-20}{39}+\frac{x-20}{49}\)
\(\Rightarrow\frac{x-20}{19}+\frac{x-20}{29}-\frac{x-20}{39}-\frac{x-20}{49}=0\)
\(\Rightarrow\left(x-20\right)\left(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\right)=0\)
vì \(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\ne0\Rightarrow x-20=0\Rightarrow x=20\)
vậy x=20
\(\frac{x}{8}=\frac{2}{x}\Rightarrow x^2=8.2\)
\(x^2=16\Rightarrow x^2=\left(\pm4\right)^2\Rightarrow x=4;x=-4\)
Học tốt
giải phương trình \(\frac{29-x}{21}\)+\(\frac{27-x}{3}\)+\(\frac{25-x}{25}\)+\(\frac{21-x}{29}\)= -5
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)
\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)
\(=\frac{1.9}{2.8}=\frac{9}{16}\)
\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-59}{39}+\frac{x-69}{49}\)
Công mỗi vế cho 2 ta được:
\(\frac{x-39}{19}+\frac{x-49}{29}+2=\frac{x-59}{39}+\frac{x-69}{49}+2\)
\(\frac{x-39}{19}+1+\frac{x-49}{29}+1=\frac{x-59}{39}+1+\frac{x-69}{49}+1\)
\(\frac{x-39}{19}+\frac{19}{19}+\frac{x-49}{29}+\frac{29}{29}=\frac{x-59}{39}+\frac{39}{39}+\frac{x-69}{49}+\frac{49}{49}\)
\(\frac{x-20}{19}+\frac{x-20}{29}=\frac{x-20}{39}+\frac{x-20}{49}\)
\(\frac{x-20}{19}+\frac{x-20}{29}-\frac{x-20}{39}-\frac{x-20}{49}=0\)
\(\left(x-20\right).\frac{1}{19}+\left(x-20\right).\frac{1}{29}-\left(x-20\right).\frac{1}{39}-\left(x-20\right).\frac{1}{49}=0\)
Đặt thừa số chung (x-20) ra ngoài
\(\left(x-20\right).\left(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\right)=0\)
\(\text{Vì }\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\ne0\text{ nên }x-20=0\Rightarrow x=20\)
mk muốn xem bài của mk đúng hay sai thôi !
chứ làm thì mk làm xong rồi !
19/41 < 21/41 , 23/53 < 23/49 và 29/61 < 33/65
Suy ra: 19/41 + 23/53 + 29/61 <21/41+ 23/49+ 33/65
Vậy A<B
Ở phép so sánh thứ 3 bạn áp dụng công thức a/b < a+n/b+n với a/b <1 và n là số tự nhiên khác 0.
Chúc bạn học tốt.
Ta có:
\(\frac{19}{41}< \frac{21}{41}\)
\(\frac{23}{53}< \frac{23}{49}\)
\(\Rightarrow\frac{19}{41}+\frac{23}{53}< \frac{21}{41}+\frac{23}{49}\)
Ta có: \(\hept{\begin{cases}\frac{29}{61}=1-\frac{32}{61}\\\frac{33}{65}=1-\frac{32}{65}\end{cases}}\)
Mà \(\frac{32}{61}>\frac{32}{65}\Rightarrow1-\frac{32}{61}< 1-\frac{32}{65}\Rightarrow\frac{29}{61}< \frac{33}{65}\)
\(\Rightarrow\frac{19}{41}+\frac{23}{53}+\frac{29}{61}< \frac{21}{41}+\frac{23}{49}+\frac{33}{65}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
Tham khảo nhé~
= 25/49 *( 21/29-7/29) +24/49 * 15/29
=25/49*14/29+ 24/49*15/29
BẠN TỰ LÀM TIẾP NHA
study well
\(\frac{25}{49}.\frac{21}{29}-\frac{25}{49}.\frac{7}{29}+\frac{24}{49}.\frac{15}{29}\)
\(=\frac{25}{49}.\left(\frac{21}{29}-\frac{7}{29}\right)+\frac{24}{49}.\frac{15}{29}\)
\(=\frac{25}{49}.\frac{14}{29}+\frac{24}{49}.\frac{15}{29}\)
\(=\left(\frac{25}{49}+\frac{24}{49}\right).\left(\frac{14}{29}+\frac{15}{29}\right)\)
\(=1.1=1\)