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Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x+y}{4+5}=\dfrac{18}{9}=2\)
Do đó: x=8; y=10
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a: \(\dfrac{x}{6}=\dfrac{8}{3}\)
=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)
b: \(\dfrac{5}{x}=\dfrac{4}{9}\)
=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)
c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)
=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)
=>x=-1-3=-4
d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)
=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)
=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)
=>\(x=-\dfrac{69}{8}\)
f: ĐKXĐ: x<>1
\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)
=>\(\left(x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)
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Áp dụng dãy tỉ số bằng nhau:
b.
\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=-5.\left(-1\right)=5\end{matrix}\right.\)
d.
\(\dfrac{4}{x}=\dfrac{7}{y}\Rightarrow\dfrac{y}{7}=\dfrac{x}{4}=\dfrac{y-x}{7-4}=\dfrac{-12}{3}=-4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.\left(-4\right)=-16\\y=7.\left(-4\right)=-28\end{matrix}\right.\)
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\(\dfrac{3}{x-2}=\dfrac{-2}{x-4}\left(dk:x\ne2;x\ne4\right)\)
\(\Rightarrow3\cdot\left(x-4\right)=-2\cdot\left(x-2\right)\)
\(\Rightarrow3x-12=-2x+4\)
\(\Rightarrow3x+2x=4+12\)
\(\Rightarrow5x=16\)
\(\Rightarrow x=\dfrac{16}{5}\left(tm\right)\)
\(ĐK:x\ne2;x\ne4\\ Có:\dfrac{3}{x-2}=\dfrac{-2}{x-4}\\ \Leftrightarrow3\left(x-4\right)=-2\left(x-2\right)\\ \Leftrightarrow3x-12=-2x+4\\ \Leftrightarrow3x+2x=4+12\\ \Leftrightarrow5x=16\\ \Leftrightarrow x=\dfrac{16}{5}\left(TM\right)\\ Vậy:x=\dfrac{16}{5}\)
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\(=\dfrac{6x^4-2x^3+5x^2-2}{3x^2-x+1}\)
\(=\dfrac{6x^4-2x^3+2x^2+3x^2-x+1+x-3}{3x^2-x+1}\)
\(=2x^2+1+\dfrac{x-3}{3x^2-x+1}\)
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\(\dfrac{7}{2}x+\dfrac{3}{4}=-\dfrac{3}{10}\\ \dfrac{7}{2}x=-\dfrac{3}{10}-\dfrac{3}{4}\\ \dfrac{7}{2}x=-\dfrac{21}{20}\\ x=-\dfrac{21}{20}:\dfrac{7}{2}\\ x=-\dfrac{21}{20}\cdot\dfrac{2}{7}\\ x=-\dfrac{3}{10}\)
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1: (3x+2)(x+2)(2x-1)
=(3x^2+6x+2x+4)(2x-1)
=(3x^2+8x+4)(2x-1)
=6x^3-3x^2+16x^2-8x+8x-4
=6x^3+13x^2-4
2: (5x+1)(x-1)+3x(2x+2)
=5x^2-5x+x-1+6x^2+6x
=11x^2+10x-1
3: 4x(2x+1)(x-1)+(x+5)(x-3)
=4x(2x^2-2x+x-1)+x^2+2x-15
=8x^3-4x^2-4x+x^2+2x-15
=8x^3-3x^2-2x-15
4: (2x-1)(x+2)(x-2)+(3x-1)(x-1)
=(2x-1)(x^2-4)+3x^2-4x+1
=2x^3-8x-x^2+4+3x^2-4x+1
=2x^3+2x^2-12x+5
vì /x-4/ luôn luôn lớn hơn 0 mà 2,6 lớn hơn 0 suy ra x-4=2,6 suy ra x =2,6+4=6,6 đây là mình hướng dẫn lập luận và cách làm khi viết trình bày thì nhớ ghi bằng kí hiệu nhé
phá dấu giá trị tuyệt đối vì 2,6 lớn hơn 0 rồi tìm x nhé bạn