Giải BPT tích:
\(\dfrac{x+2}{x+1}\le\dfrac{x-2}{x-1};\dfrac{x+5}{\left(x-7\right)\left(3-4x\right)}< 0\)
Những câu hỏi liên quan
26 tháng 2 2021 lúc 16:06
Giải bpt
\(\dfrac{1}{\sqrt{x^2+1}}+\dfrac{1}{\sqrt{3x^2+5}}\le\dfrac{2}{\sqrt{x^2-2}+1}\)
ĐKXĐ: \(x^2\ge2\)
Đặt \(\sqrt{x^2-2}=a\ge0\)
BPT tương đương: \(\dfrac{1}{\sqrt{a^2+3}}+\dfrac{1}{\sqrt{3a^2+11}}\le\dfrac{2}{a+1}\)
Ta có: \(VT^2\le2\left(\dfrac{1}{a^2+3}+\dfrac{1}{3a^2+11}\right)< 2\left(\dfrac{1}{a^2+3}+\dfrac{1}{3a^2+1}\right)=\dfrac{8\left(a^2+1\right)}{\left(3a^2+1\right)\left(a^2+3\right)}\)
Mặt khác ta có: \(\left(a-1\right)^4\ge0\Leftrightarrow a^4-4a^3+6a^2-4a+1\ge0\)
\(\Leftrightarrow3a^4+10a^2+3\ge2a^4+4a^3+4a^2+4a+2\)
\(\Leftrightarrow\left(3a^2+1\right)\left(a^2+3\right)\ge2\left(a^2+1\right)\left(a+1\right)^2\)
\(\Rightarrow\dfrac{8\left(a^2+1\right)}{\left(3a^2+1\right)\left(a^2+3\right)}\le\dfrac{4}{\left(a+1\right)^2}\)
\(\Rightarrow VT^2< \dfrac{4}{\left(a+1\right)^2}\Rightarrow VT< \dfrac{2}{a+1}\)
\(\Rightarrow\) BPT đã cho đúng với mọi \(a\ge0\) hay nghiệm của BPT là \(x^2\ge2\)
Đúng 1
Bình luận (0)
Giải BPT: \(\sqrt{x^4+x^2+1}+\sqrt{x.\left(x^2-x+1\right)}\le\sqrt{\dfrac{\left(x^2+1\right)^3}{x}}\)
Giải BPT: \(\sqrt{x^4+x^2+1}+\sqrt{x.\left(x^2-x+1\right)}\le\sqrt{\dfrac{\left(x^2+1\right)^3}{x}}\)
Giải BPT: \(\sqrt{x^4+x^2+1}+\sqrt{x.\left(x^2-x+1\right)}\le\sqrt{\dfrac{\left(x^2+1\right)^3}{x}}\)
Giai cac bpt sau
a,\(\dfrac{4x-2}{3}-x+3\le\dfrac{1-5x}{4}\)
b,\(\dfrac{x+4}{5}-x-5\ge\dfrac{x+3}{3}-\dfrac{x-2}{2}\)
a: \(\Leftrightarrow4\left(4x-2\right)+12\left(-x+3\right)< =3\left(1-5x\right)\)
=>16x-8-12x+36<=3-15x
=>4x+28<=3-15x
=>19x<=-25
hay x<=-25/19
b: \(\Leftrightarrow6\left(x+4\right)+30\left(-x-5\right)>=10\left(x+3\right)-15\left(x-2\right)\)
=>6x+24-30x-150<=10x+30-15x+30
=>-24x-126<=-5x+60
=>-19x<=186
hay x>=-186/19
\(a,\dfrac{4x-2}{3}-x+3\le\dfrac{1-5x}{4}\\ \Leftrightarrow\dfrac{4\left(4x-2\right)}{12}-\dfrac{12\left(x-3\right)}{12}\le\dfrac{3\left(1-5x\right)}{12}\\ \Leftrightarrow16x-8-12x+36\le3-15x\\ \Leftrightarrow4x+28\le3-15x\\ \Leftrightarrow19x+25\le0\\ \Leftrightarrow x\le-\dfrac{25}{19}\)
\(b,\dfrac{x+4}{5}-x-5\ge\dfrac{x+3}{3}-\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x+5\right)}{30}\ge\dfrac{10\left(x+3\right)}{30}-\dfrac{15\left(x-2\right)}{30}\\ \Leftrightarrow6x+24-30x-150\ge10x+30-15x+30\\ \Leftrightarrow-24x-126\ge-5x+60\\ \Leftrightarrow19x+186\le0\\ \Leftrightarrow x\le-\dfrac{186}{19}\)
Đúng 2
Bình luận (0)
Giari các bpt sau
a,\(\left(x+2\right)^2+3\left(x+1\right)^2\ge4x^2-4\)
b,\(\dfrac{x-1}{2}-\dfrac{x-2}{3}\le x-\dfrac{x-3}{4}\)
a: \(\Leftrightarrow x^2+4x+4+3x^2+6x+3>=4x^2-4\)
=>10x+7>=-4
=>10x>=-11
hay x>=-11/10
b: \(\Leftrightarrow6\left(x-1\right)-4\left(x-2\right)\le12x-3\left(x-3\right)\)
=>6x-6-4x+8<=12x-3x+9
=>2x+2<=9x+9
=>-7x<=7
hay x>=-1
Đúng 4
Bình luận (0)
a: ⇔x2+4x+4+3x2+6x+3>=4x2−4⇔x2+4x+4+3x2+6x+3>=4x2−4
=>10x+7>=-4
=>10x>=-11
hay x>=-11/10
b: ⇔6(x−1)−4(x−2)≤12x−3(x−3)⇔6(x−1)−4(x−2)≤12x−3(x−3)
=>6x-6-4x+8<=12x-3x+9
=>2x+2<=9x+9
=>-7x<=7
hay x>=-1
Đúng 1
Bình luận (0)
Giải BPT sau
\(\sqrt{x+1}\le\dfrac{x^2-x-2\sqrt[3]{2x+1}}{\sqrt[3]{2x+1}-3}\)
giúp mình giải bpt vs
\(\dfrac{\left|2x-1\right|-x}{2x}>1;\dfrac{2-\left|x-2\right|}{x^2-1}\ge0;\dfrac{\sqrt{x+4}-2}{4-9x^2}\le0;\dfrac{x^2-2x-3}{\sqrt[3]{3x-1}+\sqrt[3]{4-5x}}\ge0;\)\(3x^2-10x+3\ge0;\left(\sqrt{2}-x\right)\left(x^2-2\right)\left(2x-4\right)< 0;\dfrac{1}{x+9}-\dfrac{1}{x}>\dfrac{1}{2};\dfrac{2}{1-2x}\le\dfrac{3}{x+1}\)
giải bpt
\(\dfrac{1}{x^2+x+1}>\dfrac{2}{x^2+x+2}\)
\(\Leftrightarrow\dfrac{x^2+x+2-2x^2-2x-2}{\left(x^2+x+1\right)\left(x^2+x+2\right)}>0\)
=>-x^2-x>0
=>x(x-1)<0
=>0<x<1
Đúng 1
Bình luận (0)