Giải Bpt
\(4\left(x+1\right)^2< \left(2x+1\right)\left(1-\sqrt{3+2x}\right)^2\)
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giải bpt
\(\left(\sqrt{x+4}-1\right)\sqrt{x+2}\ge\frac{x^3+4x^2+3x-2\left(x+3\right)\sqrt[3]{2x+3}}{\left(\sqrt[3]{2x+3}-3\right)\left(\sqrt{x+4}+1\right)}\)
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 pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
Giải BPT\(\left(\sqrt{x+3}-\sqrt{x-1}\right)\left(\sqrt{x^2+2x+3}-2\right)\ge4\)
Giải bpt
\(\left(x-2\right)^2\ge\left(\sqrt{x-1}-1\right)^2\left(2x-1\right)\)
https://i.imgur.com/cXxRxhv.jpg
Giai cac bpt sau
a,\(\left(x+1\right)\left(2x-2\right)-3>-5x-\left(2x+1\right)\left(3-x\right)\)
b,\(\left(x-3^{ }\right)^2+4\left(2-x\right)>\left(x+7\right)\)
a: \(\Leftrightarrow2x^2-2-3>-5x+\left(2x+1\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2-5>-5x+2x^2-6x+x-3\)
\(\Leftrightarrow2x^2-5>2x^2-10x-3\)
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: \(\Leftrightarrow x^2-6x+9+8-4x>x+7\)
\(\Leftrightarrow x^2-10x+17-x-7>0\)
\(\Leftrightarrow x^2-11x+10>0\)
=>x>10 hoặc x<1
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a: ⇔2x2−2−3>−5x+(2x+1)(x−3)⇔2x2−2−3>−5x+(2x+1)(x−3)
⇔2x2−5>−5x+2x2−6x+x−3⇔2x2−5>−5x+2x2−6x+x−3
⇔2x2−5>2x2−10x−3⇔2x2−5>2x2−10x−3
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: ⇔x2−6x+9+8−4x>x+7⇔x2−6x+9+8−4x>x+7
⇔x2−10x+17−x−7>0⇔x2−10x+17−x−7>0
⇔x2−11x+10>0⇔x2−11x+10>0
=>x>10 hoặc x<1
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Giải bpt:
a,\(\frac{\sqrt{x^2-x+4}-2x-3}{x-2}>3\)
b, \(\sqrt{x\left(x-1\right)}+\sqrt{x\left(x+2\right)}\le\sqrt{x\left(4x+1\right)}\)
B1:Giải bpt sau:left(sqrt{13}-sqrt{2x^2-2x+5}-sqrt{2x^2-4x+4}right).left(x^6-x^3+x^2-x+1right)ge0B2:Cho a;b;c0 thỏa mãn a+b+cfrac{1}{a}+frac{1}{b}+frac{1}{c}.CMR 3left(a+b+cright)gesqrt{8a^2+1}+sqrt{8b^2+1}+sqrt{8c^2+1}B3:giải pt nghiệm nguyên sau : 6left(y^2-1right)+3left(x^2+y^2z^2right)+2left(z^2-9xright)0
Đọc tiếp
B1:Giải bpt sau:\(\left(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\right).\left(x^6-x^3+x^2-x+1\right)\ge0\)
B2:Cho a;b;c>0 thỏa mãn \(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\).CMR \(3\left(a+b+c\right)\ge\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\)
B3:giải pt nghiệm nguyên sau : \(6\left(y^2-1\right)+3\left(x^2+y^2z^2\right)+2\left(z^2-9x\right)=0\)
bài 2
ta có \(\left(\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\right)^2\)
\(=\left(\sqrt{a}.\sqrt{\frac{8a^2+1}{a}}+\sqrt{b}.\sqrt{\frac{8b^2+1}{b}}+\sqrt{c}.\sqrt{\frac{8c^2+1}{c}}\right)^2\)\(=\left(A\right)\)
Áp dụng bất đẳng thức Bunhiacopxki ta có;
\(\left(A\right)\le\left(a+b+c\right)\left(8a+\frac{1}{a}+8b+\frac{1}{b}+8c+\frac{8}{c}\right)\)
\(=\left(a+b+c\right)\left(9a+9b+9c\right)=9\left(a+b+c\right)^2\)
\(\Rightarrow3\left(a+b+c\right)\ge\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\)(đpcm)
Dấu \(=\)xảy ra khi \(a=b=c=1\)
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câu 1 dễ mà liên hợp đi x=\(\frac{4}{5}\)
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câu hình
ad bđt svacso
\(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_b}\ge\frac{9}{h_a+2h_b}\)
tt vs mấy cái còn lại rồi dùng S=p.r
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1) giải bpt:
\(-4\sqrt{\left(4-x\right)\left(2+x\right)}\le x^2-2x-12\)