Bất phương trình bậc nhất một ẩn - Hỏi đáp

2|x-1|=6-|x-1|

⇔ 2|x-1|+|x-1|=6

⇔ 3|x-1|=6

⇔ |x-1|=2

*ta có |x-1| = x-1 thì x-1 ≥ 0 ⇔ x ≥ 1

|x-1|=-(x-1) thì x-1 <0 ⇔ x<1

th1 vs x ≥ 1

x-1=2

⇔ x=3 (tm)

th2 vs x<1

-(x-1)=2

⇔-x+1=2

⇔-x=1

⇔ x=-1 (tm)

vậy pt có tập nghiệm là S={ -1;1}

\(a,\dfrac{2x-7}{x+3}>1\)

\(\Leftrightarrow\dfrac{2x-7}{x+3}-1>0\)

\(\Leftrightarrow\dfrac{2x-7-x-3}{x+3}>0\)

\(\Leftrightarrow\dfrac{x-10}{x+3}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10>0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10< 0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>10\\x>-3\end{matrix}\right.\\\left\{{}\begin{matrix}x\backslash< 10\\x< -3\end{matrix}\right.\end{matrix}\right.\)

Vậy x ∈ ( - ∞;-3) \(\cup\) ( 10; ∞ )

\(b,\dfrac{4x+7}{5-x}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+7>0\\5-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+7< 0\\5-x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-\dfrac{7}{4}\\x< 5\end{matrix}\right.\\\left\{{}\begin{matrix}x< -\dfrac{7}{4}\\x>5\end{matrix}\right.\end{matrix}\right.\)

TH2 : vô nghiệm

Vậy bpt có nghiệm \(-\dfrac{7}{4}< x< 5\)

áp dụngBĐt cô si cho 2 số ta có

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{a^2}{b^2}.\dfrac{b^2}{c^2}}=2\dfrac{a}{c}\)

tt ta có

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\dfrac{b}{a}\); \(\dfrac{b^2}{a^2}+\dfrac{a^2}{c^2}\ge2\dfrac{b}{c}\)

cộng các BĐT trên ta có

\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)

⇔ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\) (đpcm)

Lời giải:

Mặc định đk $a,b,c\neq 0$

Áp dụng BĐT Cô-si cho các số dương ta có:

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\geq 2\sqrt{\frac{a^2}{b^2}.\frac{b^2}{c^2}}=2|\frac{a}{c}|\geq \frac{2a}{c}\)

\(\frac{a^2}{b^2}+\frac{c^2}{a^2}\geq 2\sqrt{\frac{a^2}{b^2}.\frac{c^2}{a^2}}=2|\frac{c}{b}|\geq \frac{2c}{b}\)

\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq 2\sqrt{\frac{b^2}{c^2}.\frac{c^2}{a^2}}=2|\frac{b}{a}|\geq \frac{2b}{a}\)

Cộng theo vế:

\(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\geq 2\left(\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\right)\)

\(\Leftrightarrow \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{c}{b}+\frac{b}{a}+\frac{a}{c}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c\)

áp dụng BĐT cô si cho 2 số ko âm ta có

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}\)

<=> \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\) (đpcm)

Xét hiệu:

\(\dfrac{a}{b}+\dfrac{b}{a}-2=\dfrac{a^2+b^2-2ab}{ab}=\dfrac{\left(a-b\right)^2}{ab}\ge0\) ( luôn đúng)

\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

Ta có:\(\left(a-b\right)^2\ge0\forall a,b\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow a^2+b^2\ge2ab\)

\(\Rightarrow a^2+b^2+a^2+b^2\ge2ab+a^2+b^2\)

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2=2^2=4\)

\(\Rightarrow a^2+b^2>2\left(đpcm\right)\)

\(7x+4=3x+2018\Leftrightarrow7x-3x=2018-4\)

\(\Leftrightarrow4x=2014\Leftrightarrow x=503,5\)

\(7x+4=3x+2018\)

\(\Leftrightarrow7x-3x=2018-4\)

\(\Leftrightarrow4x=2014\)

\(\Leftrightarrow x=\dfrac{2014}{4}\)

\(\Leftrightarrow x=\dfrac{1007}{2}\)

Vậy pt có nghiệm là \(x=\dfrac{1007}{2}\)(bài này là giải pt phải k bn, nếu phải thì kluận ntn nhé)

\(\dfrac{7}{x+4}>0\Leftrightarrow x+4>0\Leftrightarrow x>-4\)

cx easy nx

\(\dfrac{7}{x+4}>0\)

\(\Leftrightarrow x+4>0\)

\(\Leftrightarrow x>-4\)

Vậy nghiệm \(x>-4\)

Gọi x(km) là độ dài q/đ đi với vận tốc 40km/h(0<x<100)

Khi đó:\(\dfrac{x}{40}\left(h\right)\)là thời gian đi với vận tốc 40km/h

\(\dfrac{100-x}{50}\left(h\right)\) là thời gian đi với vận tốc 50km/h

Theo đề ta có bpt:\(\dfrac{x}{40}+\dfrac{100-x}{50}\le\dfrac{9}{4}\)

\(\Leftrightarrow\dfrac{5x}{200}+\dfrac{4\left(100-x\right)}{200}\le\dfrac{450}{200}\)

\(\Leftrightarrow5x+400-4x\le450\)

\(\Leftrightarrow x\le50\left(tm\right)\)

1)a)\(A=\left|x\right|+\left|1+x\right|=\left|x\right|+\left|-1-x\right|\ge\left|x-1-x\right|=1\)

\(\Rightarrow MINA=1\)

b)MINB=3

2b)Ta có:\(\left(a^2-1\right)^2\ge0\)

\(\Leftrightarrow a^4-2a^2+1\ge0\)

\(\Leftrightarrow a^4+1\ge2a^2\)

\(\Rightarrow\dfrac{a^2}{a^4+1}\le\dfrac{a^2}{2a^2}=\dfrac{1}{2}\left(đpcm\right)\)

2.

a)Xét hiệu:

\(a^4+1-a\left(a^2+1\right)\)

\(=a^4+1-a^3-a\)

\(=a^3\left(a-1\right)-\left(a-1\right)\)

\(=\left(a^3-1\right)\left(a-1\right)\)

\(=\left(a-1\right)\left(a^2+a+1\right)\left(a-1\right)\)

\(=\left(a-1\right)^2\left(a^2+a+1\right)\)

Ta có:

\(a^2+a+1=a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{2}{4}=\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

Suy ra:

\(\left(a-1\right)^2\left(a^2+a+1\right)\ge0\)

=> \(a^4+1\ge a\left(a^2+1\right)\)

1) xét hiệu

\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}\ge0\)

<=> \(\dfrac{b\left(a+b\right)}{ab\left(a+b\right)}+\dfrac{a\left(a+b\right)}{ab\left(a+b\right)}-\dfrac{4ab}{ab\left(a+b\right)}\ge0\)

=> b(a+b)+a(a+b)-4ab ≥ 0

<=> ab+b²+a²+ab-4ab ≥ 0

<=> a² -2ab+b² ≥ 0

<=> (a-b)² ≥ 0 (luôn đúng )

=> đpcm

2)Ta có:\(\left(a-b\right)^2\ge0\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

TT\(\Rightarrow\left(b+c\right)^2\ge4bc;\left(c+a\right)^2\ge4ca\)

\(\Rightarrow\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\ge64a^2b^2c^2\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)