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Giải :
\(\frac{x+1}{x-2}=\frac{3}{4}\)
\(\Rightarrow4.\left(x-1\right)=3.\left(x-2\right)\)
\(\Rightarrow4x-4=3x-6\)
\(\Rightarrow4x-4-3x+6=0\)
\(\Rightarrow x+2=0\)
\(\Rightarrow x=-2\)Không thỏa mãn => Không có giá trị x thỏa mãn đề bài
\(\frac{2x-3}{x+1}=\frac{4}{7}\)
\(\Rightarrow7.\left(2x-3\right)=4.\left(x+1\right)\)
\(\Rightarrow14x-21-4x-4=0\)
\(\Rightarrow10x-25=0\)
\(\Rightarrow10x=25\)
\(\Rightarrow x=\frac{25}{10}=\frac{5}{2}\)
Giá trị trên thỏa mãn đầu bài
Các phần khác em làm tương tự nha
Em chuyển hỗn số thành phân số theo CT(\(a\dfrac{b}{c}=\dfrac{a.c+b}{c}\)) tính trong ngoặc trước rồi nhân phân số theo CT: (\(\dfrac{a}{b}.\dfrac{c}{d}=\dfrac{a.c}{b.d}\)) nha.
a) \(-1\frac{5}{7}.15+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{2}{3}-\frac{4}{5}+\frac{1}{7}\right)\)
\(=1\frac{5}{7}.\left(-15\right)+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{70}{105}-\frac{84}{105}+\frac{15}{105}\right)\)
\(=\left(-15\right)\left(1+\frac{5}{7}+\frac{2}{7}\right)+\left(-105\right).\frac{1}{105}\)
\(=-30-1=-31\)
b) \(\frac{2}{3}+\frac{3}{4}.\left(-\frac{4}{9}\right)\)
= \(=\frac{2}{3}+\frac{3.\left(-4\right)}{4.9}=\frac{2}{3}+\frac{-1}{3}=\frac{1}{3}\)
c) \(\left(\frac{3}{4}-0,2\right).\left(0,4-\frac{4}{5}\right)\)
\(=\left(\frac{3}{4}-\frac{1}{5}\right).\left(\frac{2}{5}-\frac{4}{5}\right)=\frac{11}{20}.\left(\frac{-2}{5}\right)=\frac{11.\left(-1\right).2}{2.10.5}=\frac{-11}{50}\)
a)A=\(x^5-\dfrac{1}{2}x+7x^3-2x+\dfrac{1}{5}x^3+3x^4-x^5+\dfrac{2}{5}x^4+15\)
=\(=\dfrac{-5}{2}x+\dfrac{36}{5}x^3+\dfrac{17}{5}x^4+15\)
b)B=\(3x^2-10+\dfrac{2}{5}x^3+7x-x^2+8+7x^2\)
\(=9x^2+\dfrac{2}{5}x^3+7x+2\)
c)C=\(\dfrac{1}{7}x-2x^4+5x+6\)
c , d giải nốt:
c) 2x + 3/6 = 7x - 13/15
=> (2x + 3) . 15 = (7x - 13) x 6
=> 30x +45 = 42x - 78
=> 30x - 42x = -78 - 45
=> -12x = -123
=> x = 41/4
d) 2-x/4 = 3x - 1/ - 3
=> (2-x) . -3 = 4. (3x - 1)
=> -6 - (-3x) = 12x - 4
(-6) + 4 = 12x - - (-3x)
=> -2 = 9x
=> x = -2/9
\(1,\)
\(a,\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)\)
\(=\dfrac{11}{125}+\left(\dfrac{-1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}\)
\(b,-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=\dfrac{-12}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=-15.\left[\dfrac{12}{7}+\dfrac{2}{7}+\left(-5\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\right]\)
\(=-15.\left[2+\left(-5\right).\dfrac{1}{105}\right]\)
\(=-15.\left(2-\dfrac{1}{21}\right)\)
\(=-15.\dfrac{41}{21}=\dfrac{-615}{21}\)
\(2,\)
\(a,\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=\dfrac{-15}{28}+\dfrac{11}{13}\)
\(\Leftrightarrow x=\dfrac{-15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}\)
\(\Leftrightarrow x=\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\left(\dfrac{5}{42}+\dfrac{-15}{28}\right)\)
\(\Leftrightarrow x=\dfrac{5}{12}\)
Vậy \(x=\dfrac{5}{12}\)
\(b,\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6=\dfrac{16}{10}=\dfrac{8}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=\dfrac{-8}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{5}-\dfrac{4}{15}=\dfrac{4}{3}\\x=\dfrac{-8}{5}-\dfrac{4}{15}=\dfrac{-28}{15}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{4}{3};\dfrac{-28}{15}\right\}\)
\(c,7^{x+2}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.\left(7^3+2\right)=345\)
\(\Leftrightarrow7^{x-1}.\left(343+2\right)=345\)
\(\Leftrightarrow7^{x-1}.345=345\)
\(\Leftrightarrow7^{x-1}=345:345=1\)
\(\Leftrightarrow x-1=0\)
\(x=0+1=1\)
Vậy \(x=1\)
1/ \(\frac{1}{3x}:\frac{2}{3}=1\)
<=> \(\frac{3}{3×2×x}=\:1\)
<=> \(\frac{1}{2x}=1\)<=> x = \(\frac{1}{2}\)
a, (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))
= (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\)) \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\) - \(\dfrac{21}{60}\))
= - \(\dfrac{3}{80}\) \(\times\) (- \(\dfrac{2}{3}\))
= \(\dfrac{1}{40}\)
b, (-1)3 + (- \(\dfrac{2}{3}\))2 : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
= -13 + \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1
= 0