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a) Ta có: \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow-2x+3+x+4=0\)
\(\Leftrightarrow-x+7=0\)
\(\Leftrightarrow-x=-7\)
hay x=7
Vậy: S={7}
b) Ta có: \(\dfrac{2+x}{5}-0.5x=\dfrac{1-2x}{4}+0.25\)
\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{0.5x\cdot20}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{20\cdot0.25}{20}\)
\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)
\(\Leftrightarrow8+4x-10x=5-10x+5\)
\(\Leftrightarrow-6x+8=-10x+10\)
\(\Leftrightarrow-6x+8+10x-10=0\)
\(\Leftrightarrow4x-2=0\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
d) Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-59}{1}+\dfrac{x-58}{2}+\dfrac{x-57}{3}\)
\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}-1=\dfrac{x-59}{1}-1+\dfrac{x-58}{2}-1+\dfrac{x-57}{3}-1\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{1}+\dfrac{x-60}{2}+\dfrac{x-60}{3}\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}\right)-\left(x-60\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
mà \(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)
nên x-60=0
hay x=60
Vậy: S={60}
\(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-4}{56}+\dfrac{x-5}{55}+\dfrac{x-6}{54}\)
\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}=\dfrac{x-4}{56}-1+\dfrac{x-5}{55}-1+\dfrac{x-6}{54}-1\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{56}+\dfrac{x-60}{55}+\dfrac{x-60}{54}\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}-\dfrac{1}{54}\right)=0\)
\(\Leftrightarrow x-60=0\)
\(\Rightarrow x=60\)
vậy \(S=\left\{60\right\}\)
(x -1)/59 -1 +(x-2)/58 -1 +(x-3)/57 -1 = (x-3)/56 -1 +(x-4)/55 -1 +(x-5)/54 -1
<=> (x-60)/59 +(x-60)/58 + (X-60)/57 -(x-60)/56 - (X-60)/55 -(X-60)/54 =0
<=> (x-60).(1/59 +1/58 +1/57 -1/56 -1/55 - 1/54)=0
vì 1/59 +1/58 +1/57 -1/56 -1/55 -1/54 <0
nên x-60 =0 <=> x=60
đề bài của bạn bi sai vì vế trái không thể bằng vế phải nếu đề đúng thì phải là :
(x-1)/59 +(x-2)/58 +(x-3)/57 =(x-4)/56 +(x-5)/55 +(x-6)/54
khí đó bạn giải cách như trên ,chúc bạn học toán tốt
Xin lỗi mình làm hơi tắt nha !!!Còn 1 cách nữa ,nếu bạn muốn thì nói với mình nha !!
Ta có : \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
\(\Leftrightarrow\frac{x}{59}+\frac{x}{58}+\frac{x}{57}-\frac{x}{56}-\frac{x}{55}-\frac{x}{54}=\frac{1}{59}+\frac{2}{58}+\frac{3}{57}-\frac{4}{56}-\frac{5}{55}-\frac{6}{54}\)
<=> x = 60
Vậy x = 60
Bạn kiểm tra lại đề nhé. Chỗ
\(.....=\frac{x-4}{56}+\frac{x-5}{56}+\frac{x-6}{54}\)
a) \(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Leftrightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)
\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)
\(\Leftrightarrow\) \(\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
Vì \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)
\(\Rightarrow x+95=0\)
\(\Leftrightarrow x=-95\)
Vậy phương trình có một nghiệm x = -95
b) \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
\(\Leftrightarrow\left(\frac{x-1}{59}-1\right)+\left(\frac{x-2}{58}-1\right)+\left(\frac{x-3}{57}-1\right)=\left(\frac{x-4}{56}-1\right)+\left(\frac{x-5}{55}-1\right)+\left(\frac{x-6}{54}-1\right)\)
\(\Leftrightarrow\frac{x-60}{59}+\frac{x-60}{58}+\frac{x-60}{57}-\frac{x-60}{56}-\frac{x-60}{55}-\frac{x-60}{54}=0\)
\(\Leftrightarrow\left(x-60\right)\left(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\right)=0\)
Vì \(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\ne0\)
\(\Rightarrow x-60=0\)
\(\Leftrightarrow x=60\)
Vậy phương trình có một nghiệm x = 60
a) \(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Rightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)
\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)
\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)
\(\Rightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
Mà \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)
\(\Rightarrow x+95=0\)
\(\Rightarrow x=-95\)
Vậy x = -95
b) \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
\(\Rightarrow\left(\frac{x-1}{59}-1\right)+\left(\frac{x-2}{58}-1\right)+\left(\frac{x-3}{57}-1\right)=\left(\frac{x-4}{56}-1\right)+\left(\frac{x-5}{55}-1\right)+\left(\frac{x-6}{54}-1\right)\)
\(\Rightarrow\frac{x-60}{59}+\frac{x-60}{58}+\frac{x-60}{57}-\frac{x-60}{56}-\frac{x-5}{55}-\frac{x-6}{54}=0\)
\(\Rightarrow\left(x-60\right)\left(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\right)=0\)
Mà \(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\ne0\)
\(\Rightarrow x-60=0\)
\(\Rightarrow x=60\)
Vậy x = 60
2) Ta có: \(\dfrac{59-x}{19}+\dfrac{58-x}{18}=\dfrac{57-x}{17}+\dfrac{56-x}{16}\)
\(\Leftrightarrow\dfrac{59-x}{19}-1+\dfrac{58-x}{18}-1=\dfrac{57-x}{17}-1=\dfrac{56-x}{16}-1\)
\(\Leftrightarrow\dfrac{40-x}{19}+\dfrac{40-x}{18}-\dfrac{40-x}{17}-\dfrac{40-x}{16}=0\)
\(\Leftrightarrow\left(40-x\right)\left(\dfrac{1}{19}+\dfrac{1}{18}-\dfrac{1}{17}-\dfrac{1}{16}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}-\dfrac{1}{17}-\dfrac{1}{16}\ne0\)
nên 40-x=0
hay x=40
Vậy: x=40
\(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)
\(< =>\dfrac{x+1}{59}+1+\dfrac{x+3}{57}+1+\dfrac{x+5}{55}+1=\dfrac{x+7}{53}+1+\dfrac{x+9}{51}+1+\dfrac{x+11}{49}+1\)
\(< =>\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}=\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)
\(< =>\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)=0\\ < =>x+60=0\\ < =>x=-60\)
Ta có : \(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)
\(\Leftrightarrow\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}+3\text{=}\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}+3\)
\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)
\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)
\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}\text{=}\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)
\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}-\dfrac{x+60}{53}-\dfrac{x+60}{51}-\dfrac{x-60}{49}\text{=}0\)
\(\Leftrightarrow\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)\text{=}0\)
\(Do\) \(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\ne0\)
\(\Leftrightarrow\left(x+60\right)\text{=}0\)
\(x\text{=}-60\)
\(Vậy...\)