Giải pt sau \(\left(\frac{x-3}{x-2}\right)^3-\left(x-3\right)^3=16\)
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B
2
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 3 2019
\(x\ne2\)
Áp dụng HĐT \(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)
\(\left(\frac{x-3}{x-2}\right)^3-\left(x-3\right)^3=16\)
\(\Leftrightarrow\left(\frac{\left(x-3\right)-\left(x-3\right)\left(x-2\right)}{x-2}\right)^3+\frac{3\left(x-3\right)^2}{\left(x-2\right)}\left(\frac{x-3}{x-2}-x+3\right)=16\)
\(\Leftrightarrow\left(\frac{\left(x-3\right)\left(3-x\right)}{\left(x-2\right)}\right)^3+\frac{3\left(x-3\right)^2}{x-2}\left(\frac{\left(x-3\right)\left(3-x\right)}{x-2}\right)=16\)
\(\Leftrightarrow\left(-\frac{\left(x-3\right)^2}{x-2}\right)^3-3.\left(\frac{\left(x-3\right)^2}{x-2}\right)^2=16\)
Đặt \(\frac{\left(x-3\right)^2}{x-2}=a\)
\(-a^3-3a^2=16\Leftrightarrow a^3+3a^2+16=0\Rightarrow a=-4\)
\(\Rightarrow\frac{\left(x-3\right)^2}{x-2}=-4\Leftrightarrow x^2-2x+1=0\Rightarrow x=1\)
H
0
TP
22 tháng 8 2018
\(A=\left(\frac{\sqrt{3}}{x^2+x\sqrt{x}+3}+\frac{3}{x^3-\sqrt{27}}\right)\left(\frac{x}{\sqrt{3}}+\frac{\sqrt{3}}{x}+1\right)\)
\(\Leftrightarrow A=\left[\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}+\frac{3}{\left(x-\sqrt{3}\right)\left(x+x\sqrt{3}+3\right)}\right]\left(\frac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\right)\)
\(\Leftrightarrow A=\frac{x\sqrt{3}-3+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}.\frac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(\Leftrightarrow A=\frac{1}{x-\sqrt{3}}\)
7 tháng 5 2020
Bài 1:
a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)
\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)
Suy ra: \(12x-45-12x^2+45x=0\)
\(\Leftrightarrow-12x^2+57x-45=0\)
\(\Leftrightarrow-12x^2+12x+45x-45=0\)
\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)
\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)
mà \(-3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)
b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)
Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)
\(\Leftrightarrow-x^2+16x-39=0\)
\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)
\(\Leftrightarrow x^2-13x-3x+39=0\)
\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)
Vậy: Tập nghiệm S={3;13}
c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)
\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)
Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)
\(\Leftrightarrow-21x^2+26x+11=0\)
\(\Leftrightarrow-21x^2-7x+33x+11=0\)
\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
11 tháng 4 2020
Hệ phương trình đề cho tương đương
\(\left\{{}\begin{matrix}\frac{1}{2}xy+18=\frac{1}{2}xy+x+y+2\\\frac{1}{2}xy-16=\frac{1}{2}xy+\frac{3}{2}x-y-3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2=18\\\frac{3}{2}x-y-3=-16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=16\\\frac{3}{2}x-y=-13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{3}{2}x=3\\x+y=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{6}{5}\\y=\frac{74}{5}\end{matrix}\right.\)
KL: ........................
30 tháng 3 2020
ĐK: \(x\in R\backslash\left\{-4,-3,-2,-1\right\}\)
PT ban đầu
\(\Leftrightarrow\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}+\frac{x+3-x-2}{\left(x+2\right)\left(x+3\right)}+\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}+\frac{x+5-x-4}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+5}=403\\ \Leftrightarrow x+5=\frac{1}{403}\Leftrightarrow x=\frac{-2014}{403}\)
Chúc bạn học tốt nha
.
30 tháng 3 2020
Sr bạn nha, nhưng điều kiện là \(x\in R\backslash\left\{-5,-4,-3,-2,-1\right\}\). (Xét thiếu :>)
Chúc bạn học tốt nha.