Tìm X
( X-\(\frac{2}{3}\)) x3 = \(\frac{3}{4}\)
Những câu hỏi liên quan
tìm x:
a) \(\overline{x3}+\overline{3x}=12\times11\)
b) \(4\frac{3}{4}-\left(\frac{1}{2}+x\right)\div4\frac{2}{3}=2\frac{1}{2}\)
Tìm các số nguyên x,y,z,t biết:
$\frac{27}{4}$274 =$\frac{-x}{3}$−x3 =$\frac{\left(z+3\right)^3}{-4}$(z+3)3−4 =$\frac{\left|t-2\right|}{8}$//t/−2/8
chú ý / là giá trị tuyệt đối
let P(x) be a polynomial of degree 3 and x1, x2, x3 are the solutions of P(x)=0. let \(\frac{P\left(\frac{1}{3}\right)-P\left(\frac{-1}{3}\right)}{P\left(0\right)}=8,\frac{P\left(\frac{1}{4}\right)-P\left(\frac{-1}{4}\right)}{P\left(0\right)}=9\)and x1+x2+x3 = 35. find the value of \(\frac{x2+x3}{x1}+\frac{x1+x3}{x2}+\frac{x1+x2}{x3}\)
Tìm m để phương trình :(x^2-1)(x+3)(x+5)=m có 4 nghiệm phân biệt thỏa mãn \(\frac{1}{x1}+\frac{1}{x2}+\frac{1}{x3}+\frac{1}{x4}=-1\)
Phân tích thành nhân tử:frac{1}{27}x^{3}-frac{1}{6}x^{2}+frac{1}{4}x-frac{1}{8}271x3−61x2+41x−81
Đọc tiếp
Phân tích thành nhân tử:
\frac{1}{27}x^{3}-\frac{1}{6}x^{2}+\frac{1}{4}x-\frac{1}{8} =
Tìm các số nguyên x,y,z,t biết:
$\frac{27}{4}$274 =$\frac{-x}{3}$−x3 =$\frac{\left(z+3\right)^3}{-4}$(z+3)3−4 =$\frac{\left|t-2\right|}{8}$//t/−2/8
ai nhanh mk tick nha
Tìm X1 ,x2 ,x3,x4,x5
\(\frac{x1-1}{5}=\frac{x2-2}{4}=\frac{x3-3}{3}=\frac{x4-4}{2}=\frac{x5-5}{1}\)\(v\text{à}\)X1 +x2+x3+x4+x5 =30
Đặt \(\frac{x_1-1}{5}=\frac{x_2-2}{4}=\frac{x_3-3}{3}=\frac{x_4-4}{2}=\frac{x_5-5}{1}=k\)
Áp dụng TC DTSBN ta có :
\(k=\frac{\left(x_1-1\right)+\left(x_2-2\right)+\left(x_3-3\right)+\left(x_4-4\right)+\left(x_5-5\right)}{5+4+3+2+1}\)
\(=\frac{x_1+x_2+x_3+x_4+x_5-15}{15}=\frac{30-15}{15}=1\)
\(\frac{x_1-1}{5}=1\Rightarrow x_1=6;\frac{x_2-2}{4}=1\Rightarrow x_2=6;\frac{x_3-3}{3}=1\Rightarrow x_3=6;\frac{x_4-4}{2}=1\Rightarrow x_4=6;\frac{x^5-5}{2}=1\Rightarrow x_5=6\)
Vậy \(x_1=x_2=x_3=x_4=x_5=6\)
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-\(\left(x-\frac{2}{3}\right)x3=\frac{3}{4}\)
MK CẦN GẤP
-\(\left(x-\frac{2}{3}\right).3=\frac{3}{4}\)
-\(\left(x-\frac{2}{3}\right)=\frac{3}{4}:3\)
-\(x-\frac{2}{3}=\frac{1}{4}\)
\(-x=\frac{1}{4}+\frac{2}{3}\)
\(-x=\frac{11}{12}\)
\(=>x=-\frac{11}{12}\)
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<br class="Apple-interchange-newline"><div id="inner-editor"></div>(x−23 ).3=34
-(x−23 )=34 :3
-x−23 =14
−x=14 +23
−x=1112
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x.2\(\frac{3}{14}\)+ x3\(\frac{2}{7}\)= 5 - 2\(\frac{3}{4}\)
x.\(2\frac{3}{14}\)+ x.\(3\frac{2}{7}\)= 5 - \(2\frac{3}{4}\)
x.\(\frac{31}{14}\)+ x.\(\frac{23}{7}\)=5 -\(\frac{11}{4}\)
x (\(\frac{31}{14}\)+ \(\frac{23}{7}\) = \(\frac{20}{4}\)- \(\frac{11}{4}\)
x ( \(\frac{31}{14}\)+ \(\frac{46}{14}\))=\(\frac{9}{4}\)
x\(\frac{11}{2}\) =\(\frac{9}{4}\)
x =\(\frac{9}{4}\):\(\frac{11}{2}\)
x =\(\frac{9}{4}\). \(\frac{2}{11}\)
x =\(\frac{9}{2.11}\)
x =\(\frac{9}{22}\)
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