tìm x :
\(x+1\frac{1}{2}=2\frac{1}{2}\)
\(x-1,2=1,1\)
Những câu hỏi liên quan
A ) / X + 2 / + / X + / + / X + frac{1}{2}/ 4X b ) / x + 1,1 / + / x + 1,2 / + / x + 1,3 / + / x + 1,4 / 5xc ) / x + frac{1}{1.3}/ + / x + frac{1}{3.5}/ + / x +frac{1}{5.7} / + ........ + / x + frac{1}{97.99}/ 50xd ) / x + frac{1}{1.5}/ + / x + frac{1}{5.9}/ + / x + frac{1}{9.13}/ + ......... + / x + frac{1}{397.401}/ 101x
Đọc tiếp
A ) / X + 2 / + / X + / + / X + \(\frac{1}{2}\)/ = 4X
b ) / x + 1,1 / + / x + 1,2 / + / x + 1,3 / + / x + 1,4 / = 5x
c ) / x + \(\frac{1}{1.3}\)/ + / x + \(\frac{1}{3.5}\)/ + / x +\(\frac{1}{5.7}\) / + ........ + / x + \(\frac{1}{97.99}\)/ = 50x
d ) / x + \(\frac{1}{1.5}\)/ + / x + \(\frac{1}{5.9}\)/ + / x + \(\frac{1}{9.13}\)/ + ......... + / x + \(\frac{1}{397.401}\)/ = 101x
Bài này khá ez thôi:
a) bạn sửa lại đề rồi làm theo cách làm của b,c,d nhé
b) Ta có: \(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|\ge0\left(\forall x\right)\)
\(\Rightarrow5x\ge0\Rightarrow x\ge0\) khi đó:
\(PT\Leftrightarrow x+1,1+x+1,2+x+1,3+x+1,4=5x\)
\(\Leftrightarrow x=5\)
c,d tương tự nhé
c,\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}+\right|+...+\left|x+\frac{1}{97.99}\right|\ge0\forall x\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)Khi đó:
\(x+\frac{1}{1.3}+x+\frac{1}{3.5}+...+x+\frac{1}{97.99}=50x\)
\(\Rightarrow49x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)=50x\)
\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{49}{99}\)
cho \(y=\frac{x^2+\frac{1}{x^2}}{x^2-\frac{1}{x^2}},z=\frac{x^4+\frac{1}{x^4}}{x^4-\frac{1}{x^4}},x\ne0,-1,1\). Hãy biểu diễn z theo y
Ta có : \(y=\frac{x^2+\frac{1}{x^2}}{x^2-\frac{1}{x^2}}=\frac{x^4+1}{x^4-1}\); \(z=\frac{x^4+\frac{1}{x^4}}{x^4-\frac{1}{x^4}}=\frac{x^8+1}{x^8-1}\)
\(y+\frac{1}{y}=\frac{x^4+1}{x^4-1}+\frac{x^4-1}{x^4+1}=\frac{\left(x^4+1\right)^2+\left(x^4-1\right)^2}{x^8-1}=\frac{2\left(x^8+1\right)}{x^8-1}=2z\)
\(\Rightarrow z=\frac{y+\frac{1}{y}}{2}=\frac{y^2+1}{2y}\)
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Tìm x biết : \(\frac{\frac{2}{3}+\frac{1}{3}\left(1,2-x\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}=\frac{1}{4}\)
BÀI 1 TÍNH :
A,\(\frac{4}{9}\)+1,2(31)-0,(13)
B,2\(\frac{1}{2}\)-3,4(12)-4/3+1/3.(\(\frac{1}{2}\)+0,5- 3\(\frac{1}{2}\))
BÀi 2 Tìm X
0,(26).x = 1,2(31)
Bài 2. Tìm x biết: \(\left\{\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{63.65}\right\}.13-\orbr{ }3,08:\left(x+1,2\right):0,7=\frac{1}{5}\)
\(\frac{x+1}{x+2}=\frac{0,8}{1,2}\)
\(\frac{1}{4}x+5-\frac{2}{5}x=1-x\)
\(\frac{x+1}{x+2}=\frac{0,8}{1,2}\)
\(\Rightarrow1,2\left(x+1\right)=0,8\left(x+2\right)\)
\(1,2x+1,2=0,8x+0,16\)
\(1,2x-0,8x=0,16-1,2\)
\(0,4x=-1,04\)
\(x=-2,6\)
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\(\frac{1}{4}x+5-\frac{2}{5}x=1-x\)
\(0,85x=-4\)
\(x=-\frac{80}{17}\)
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cho x,y,z \(\ne\)-1,1 ,xy+zx+zy=1
\(\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}=\frac{4xyz}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}\)
Tìm x biết:
a) \(\frac{1\frac{2}{3}}{3}\)= \(\frac{\frac{1}{2}}{x-\frac{1}{3}}\)
b) \(\frac{x-2}{\frac{3}{2}}\)= \(\frac{x+1}{1,2}\)
Tìm x, biết:
b)3/x+4/-/2x+1/-5/x+3/+/x-9/=5
c)\(\left|\frac{11}{5}-x\right|+\left|x-\frac{1}{5}\right|+\frac{41}{5}=1,2\)
d)\(2\left|x+\frac{7}{2}\right|+\left|x\right|-\frac{7}{2}=\left|\frac{11}{5}-x\right|\)
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