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C= x^6+27/x^4 - 3x^3 +6x^2 -9x + 9
= (x^2+3)(x^4-3x^2+9)/(x^4+3x^2)-(3x^3+9x)+(3x^2+9)
=(x^2+3)(x^4+6x^2+9-9x^2)/(x^2+3x)(x^2-3x+3)
= (x^2+3+3x)(x^2+3-3x)/x^2+3-3x =x^2+3x+3
=(x^2+3x+9/4) -9/4+3 = (x+3/2)^2 +3/4 >= 3/4
Dấu = xảy ra khi x=-3/2
Vậy Cmin = 3/4 <=> x=-3/2
a: \(A=x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)
Dấu '=' xảy ra khi x=3/2
c: \(x^2-x+2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\)
=>\(\dfrac{3}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}< =3:\dfrac{7}{4}=\dfrac{12}{7}\)
=>C>=-12/7
Dấu '=' xảy ra khi x=1/2
Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!
a:6x-5-9x^2
=-(9x^2-6x+5)
=-(9x^2-6x+1+4)
=-(3x-1)^2-4<=-4
=>A>=2/-4=-1/2
Dấu = xảy ra khi x=1/3
b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)
2x^2-3x+2=2(x^2-3/2x+1)
=2(x^2-2*x*3/4+9/16+7/16)
=2(x-3/4)^2+7/8>=7/8
=>-1/2x^2-3x+2<=-1:7/8=-8/7
=>B<=-8/7+2=6/7
Dâu = xảy ra khi x=3/4
`a,` \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
`<=> (5(5x+2))/30 - (10(8x-1))/30 = (6(4x+2))/30 - (5.30)/30`
`<=> 5(5x+2) - 10(8x-1) =6(4x+2) - 5.30`
`<=> 25x + 10 - 80x + 10 = 24x+12 - 150`
`<=> -55x +20 = 24x-138`
`<=> -55x -24x=-138-20`
`<=>-79x=-158`
`<=> x=2`
Vậy pt có nghiệm `x=2`
`b,` \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)
Ta có : `(x+2)/(x-2) -1/x = 2/(x(x-2))`
`<=> (x(x+2))/(x(x-2)) - (x-2)/(x(x-2)) = 2/(x(x-2))`
`=> x^2 +2x - x +2 = 2`
`<=> x^2 + x =0`
`<=>x(x+1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-1\end{matrix}\right.\)
Vậy pt có nghiệm `x=-1`
`c,2x^3 + 6x^2 =x^2 +3x`
`<=> 2x^3 + 6x^2 -x^2 -3x=0`
`<=> 2x^3 + 5x^2 -3x=0`
`->` Đề có sai ko ạ ?
`d,` \(\left|x-4\right|+3x=5\) `(1)`
Thường hợp `1` : `x-4 >= 0<=> x >=0` thì phương trình `(1)` thở thành :
`x-4 = 5-3x`
`<=> x+3x=5+4`
`<=> 4x=9`
`<=> x= 9/4 (t//m)`
Trường hợp `2` : `x-4< 0<=> x<0` thì phương trình `(1)` trở thành :
`-(x-4) =5-3x`
`<=> -x +4=5-3x`
`<=> -x+3x=5-4`
`<=> 2x =1`
`<=>x=1/2 ( kt//m)`
Vậy phương trình có nghiệm `x=9/4`
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a: \(x-3\left(2x-6\right)=21-\left(5x+3\right)\)
=>\(x-6x+18=21-5x-3\)
=>18=18(luôn đúng)
=>\(x\in R\)
b: \(\left(x-2\right)\left(x+2\right)-\left(x-1\right)^2=2\left(x+1\right)\)
=>\(x^2-4-x^2+2x-1=2x+2\)
=>2x-5=2x+2
=>-7=0(vô lý)
=>\(x\in\varnothing\)
c: \(\dfrac{9x+4}{6}=1-\dfrac{3x-5}{9}\)
=>\(\dfrac{3\left(9x+4\right)}{18}=\dfrac{18}{18}-\dfrac{2\left(3x-5\right)}{18}\)
=>3(9x+4)=18-2(3x-5)
=>27x+12=18-6x+10
=>27x+12=-6x+28
=>33x=16
=>\(x=\dfrac{16}{33}\left(nhận\right)\)
d: ĐKXĐ: \(x\notin\left\{2;5\right\}\)
\(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
=>\(\dfrac{6x+1}{\left(x-2\right)\left(x-5\right)}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
=>\(6x+1+5\left(x-5\right)=3\left(x-2\right)\)
=>6x+1+5x-25=3x-6
=>11x-24=3x-6
=>8x=18
=>\(x=\dfrac{9}{4}\left(nhận\right)\)
a: x−3(2x−6)=21−(5x+3)
=>x−6x+18=21−5x−3
=>18=18(luôn đúng)
=>x∈R
b: (x−2)(x+2)−(x−1)2=2(x+1)
=>x2−4−x2+2x−1=2x+2
=>2x-5=2x+2
=>-7=0(vô lý)
=>x∈∅
c: 9x+46=1−3x−59
=>3(9x+4)18=1818−2(3x−5)18
=>3(9x+4)=18-2(3x-5)
=>27x+12=18-6x+10
=>27x+12=-6x+28
=>33x=16
=>x=1633(nhận)
d: ĐKXĐ: x∉{2;5}
6x+1x2−7x+10+5x−2=3x−5
=>6x+1(x−2)(x−5)+5x−2=3x−5
=>6x+1+5(x−5)=3(x−2)6
=>6x+1+5x-25=3x-6
=>11x-24=3x-6
=>8x=18
=>x=94(nhận)
\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)
= \(\dfrac{3x^2-6x+15+2}{x^2-2x+5}\)
=\(\dfrac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}\)
= \(\dfrac{3\cdot\left(x^2-2x+5\right)}{x^2-2x+5}+\dfrac{2}{x^2-2x+5}\)
= \(3+\dfrac{2}{x^2-2x+5}\)
= \(3+\dfrac{2}{x^2-2x+1+4}\)
= \(3+\dfrac{2}{\left(x-1\right)^2+4}\)
vì (x-1)2 ≥ 0 ∀ x
⇔ (x-1)2 +4 ≥ 4
⇔\(\dfrac{2}{\left(x-1\right)^2+4}\le\dfrac{1}{2}\)
⇔\(3+\dfrac{2}{\left(x-1\right)^2+4}\le\dfrac{7}{2}\)
⇔ A \(\le\dfrac{7}{2}\)
⇔ Min A =\(\dfrac{7}{2}\)
khi x-1=0
⇔ x=1
vậy ....
Ta có:\(B=\dfrac{2x^2-16x+41}{x^2-8x+22}\)
\(B=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)
\(B=2-\dfrac{3}{x^2-8x+16+6}\)
\(B=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{5}{2}\)
\(\Rightarrow MINB=\dfrac{5}{2}\Leftrightarrow x=4\)