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a: \(A=\left|3x-9\right|+1.5\ge1.5\forall x\)
Dấu '=' xảy ra khi x=3
b: \(B=\left|x-7\right|-14\ge-14\forall x\)
Dấu '=' xảy ra khi x=7
a) \(2x\left(3x+1\right)+3x\left(4-2x\right)=7\)
\(\Rightarrow6x^2+2x+12x-6x^2=7\)
\(\Rightarrow14x=7\Rightarrow x=\frac{1}{2}\)
b) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(-80x=-480\)
x = 6
c) \(\left(3x+2\right).\left(2x+9\right)-\left(x+2\right).\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Rightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x+1-x+6\) ( chỗ này bn tự phân tích ik nha, mk chỉ đưa ra kp sau khi phân tích thôi, ko thì viết ra dài lắm)
\(\Rightarrow18x+16=7\)
18x = -9
x = -2
18x =
1.
a) \(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)
⇒ \(\left(x-2\right).\left(x+7\right)=\left(x+4\right).\left(x-1\right)\)
⇒ \(x^2+7x-2x-14=x^2-x+4x-4\)
⇒ \(x^2+7x-2x-14-x^2+x-4x+4=0\)
⇒ \(2x-10=0\)
⇒ \(2x=0+10\)
⇒ \(2x=10\)
⇒ \(x=10:2\)
⇒ \(x=5\)
Vậy \(x=5.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!
a: \(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
=>14x=7
hay x=1/2
b: \(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)
=>-56x+156=24x-324
=>-80x=-480
hay x=6
c: \(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6=7\)
=>18x+16=7
=>18x=-9
hay x=-1/2
\(/x-\frac{1}{2}/=\frac{1}{3}\\ =>\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{1}{3}+\frac{1}{2}\\x=-\frac{1}{3}+\frac{1}{2}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)
\(a,|x-\frac{1}{2}|=\frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}}\)
\(b,\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\)
\(\frac{28}{27}=x:\frac{3}{7}\)
\(x=\frac{4}{9}\)
a) \(\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\Rightarrow\frac{28}{27}=x:\frac{3}{7}\Rightarrow x=\frac{4}{9}\)
b) \(\left(x-\frac{4}{7}\right)^3=343\Rightarrow\left(x-\frac{4}{7}\right)^3=7^3\Rightarrow x-\frac{4}{7}=7\Rightarrow x=\frac{53}{7}\)
c) \(x^5=x^3\Leftrightarrow\hept{\begin{cases}x=1\\x=0\end{cases}}\)
e) \(\left(x-1\right)^4=16\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^4=2^4\\\left(x-1\right)^4=\left(-2\right)^4\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
\(\frac{4}{7}x-x=-\frac{9}{14}\)
\(\Rightarrow\left(\frac{4}{7}-1\right)x=-\frac{9}{14}\)
\(\Rightarrow-\frac{3}{7}x=-\frac{9}{14}\)
\(\Rightarrow x=\frac{9}{14}.\frac{7}{3}=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{2}\)