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|9x−8|+|7x−6|+|5x−4|+|3x−2|+x=0(1)|9x−8|+|7x−6|+|5x−4|+|3x−2|+x=0(1).
Vì |9x−8|+|7x−6|+|5x−4|+|3x−2|>0∀x|9x−8|+|7x−6|+|5x−4|+|3x−2|>0∀x
Nên từ (1) ⇒x<0⇒9x−8;7x−6;5x−4;3x−2<0⇒x<0⇒9x−8;7x−6;5x−4;3x−2<0.
Phương trình (1) trở thành:
8−9x+6−7x+4−5x+2−3x+x=0⇔20−23x=0⇔x=20/23>0(ktm)
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1/ 0, 71
2/ Tương tự 2 câu 1, 3 nhé!
3/ 11,25
Tick đúng nha! Thanks!
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Answer:
3.
\(x^2+2y^2+2xy+7x+7y+10=0\)
\(\Rightarrow\left(x^2+2xy+y^2\right)+7x+7y+y^2+10=0\)
\(\Rightarrow\left(x+y\right)^2+7.\left(x+y\right)+y^2+10=0\)
\(\Rightarrow4S^2+28S+4y^2+40=0\)
\(\Rightarrow4S^2+28S+49+4y^2-9=0\)
\(\Rightarrow\left(2S+7\right)^2=9-4y^2\le9\left(1\right)\)
\(\Rightarrow-3\le2S+7\le3\)
\(\Rightarrow-10\le2S\le-4\)
\(\Rightarrow-5\le S\le-2\left(2\right)\)
Dấu " = " xảy ra khi: \(\left(1\right)\Rightarrow y=0\)
Vậy giá trị nhỏ nhất của \(S=x+y=-5\Rightarrow\hept{\begin{cases}y=0\\x=-5\end{cases}}\)
Vậy giá trị lớn nhất của \(S=x+y=-2\Rightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}\)
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\(\dfrac{x}{x^2+x+1}=\dfrac{1}{4}\)
=>\(x^2+x+1=4x\)
=>\(x^2-3x+1=0\)
\(\dfrac{x^5-3x^3-10x+12}{x^4+7x^2+15}\)
\(=\dfrac{x^5-3x^4+x^3+3x^4-9x^3+3x^2+5x^3-15x^2+5x+12x^2-36x+12+21x}{x^4+7x^2+15}\)
\(=\dfrac{x^3\left(x^2-3x+1\right)+3x^2\left(x^2-3x+1\right)+5x\left(x^2-3x+1\right)+12\left(x^2-3x+1\right)+21x}{x^4+7x^2+15}\)
\(=\dfrac{21x}{x^4-3x^3+x^2+3x^3-9x^2+3x+15x^2-45x+15+42x}\)
\(=\dfrac{21x}{x^2\left(x^2-3x+1\right)+3x\left(x^2-3x+1\right)+15\left(x^2-3x+1\right)+42x}\)
\(=\dfrac{21x}{42x}=\dfrac{1}{2}\)
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\(\Rightarrow7x^3+7x^2-4x^2-4x+x+1=0\\ \Rightarrow\left(x+1\right)\left(7x^2-4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\7x^2-4x+1=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Rightarrow7\left(x^2-2\cdot\dfrac{2}{7}x+\dfrac{4}{49}\right)+\dfrac{3}{7}=0\\ \Rightarrow7\left(x-\dfrac{2}{7}\right)^2+\dfrac{3}{7}=0\left(\text{vô lí}\right)\)
Vậy x=-1
Answer:
\(7x^3+3x^2-3x+1=0\)
\(\Rightarrow7x^3+7x^2-4x^2-4x+x+1=0\)
\(\Rightarrow7x^2.\left(x+1\right)-4x.\left(x+1\right)+\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right).\left(7x^2-4x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\7x^2-4x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\7x^2-4x+1=0\text{(Vô lý)}\end{cases}}\)