giải các BPT
1. \(\frac{1-\sqrt{1-4x^2}}{x}< 3\)
2.\(\sqrt[3]{2-x}+\sqrt{x-1}>1\)
3.\(x+\frac{x}{\sqrt{x^2-1}}>\frac{35}{12}\)
giải bpt
1.\(\sqrt{5x+1}-\sqrt{4x-1}\le3\sqrt{x}\)
2.\(\frac{\sqrt{2\left(x^2-16\right)}}{\sqrt{x-3}}+\sqrt{x-3}>\frac{7-x}{\sqrt{x-}}\)
ĐKXĐ: \(x\ge\frac{1}{4}\)
\(\sqrt{5x+1}\le3\sqrt{x}+\sqrt{4x-1}\)
\(\Leftrightarrow5x+1\le9x+4x-1+6\sqrt{4x^2-x}\)
\(\Leftrightarrow3\sqrt{4x^2-x}\ge1-4x\)
Do \(x\ge1\Rightarrow\left\{{}\begin{matrix}1-4x\le0\\\sqrt{4x^2-x}\ge0\end{matrix}\right.\) \(\Rightarrow\) BPT luôn đúng
Vậy nghiệm của BPT là \(x\ge\frac{1}{4}\)
b/ ĐKXĐ: \(x\ge4\)
\(\Leftrightarrow\sqrt{2\left(x^2-16\right)}+x-3>7-x\)
\(\Leftrightarrow\sqrt{2\left(x^2-16\right)}>10-2x\)
- Với \(x>5\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT luôn đúng
- Với \(x\le5\) bình phương 2 vế:
\(2\left(x^2-16\right)>4\left(x-5\right)^2\)
\(\Leftrightarrow x^2-20x+66< 0\)
\(\Rightarrow10-\sqrt{34}< x< 10+\sqrt{34}\)
Vậy nghiệm của BPT là \(x>10-\sqrt{34}\)
Giải bpt
a) \(\frac{3}{\sqrt{x-2}-1}\ge\frac{5}{\sqrt{x-2}-3}\)
b) \(x\sqrt{x-3}-\frac{\sqrt{x-3}}{2-x}\le0\)
c) \(\frac{2\sqrt{x-1}-4}{\sqrt{4-x^2}-1}\ge2-\sqrt{x-1}\)
a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\x\ne\left\{3;11\right\}\end{matrix}\right.\)
Đặt \(\sqrt{x-2}=t\ge0\)
\(\Rightarrow\frac{3}{t-1}\ge\frac{5}{t-3}\)
\(\Leftrightarrow\frac{3}{t-1}-\frac{5}{t-3}\ge0\)
\(\Leftrightarrow\frac{3t-9-5t+5}{\left(t-1\right)\left(t-3\right)}\ge0\)
\(\Leftrightarrow\frac{-2t-4}{\left(t-1\right)\left(t-3\right)}\ge0\)
\(\Leftrightarrow\frac{t+2}{\left(t-1\right)\left(t-3\right)}\le0\)
\(\Leftrightarrow1< t< 3\)
\(\Rightarrow1< \sqrt{x-2}< 3\)
\(\Leftrightarrow1< x-2< 9\Rightarrow3< x< 11\)
b/
ĐKXĐ: \(x\ge3\)
- Với \(x=3\) BPT thỏa mãn
- Với \(x>3\Rightarrow\sqrt{x-3}>0\) BPT tương đương
\(x-\frac{1}{2-x}\le0\Leftrightarrow x+\frac{1}{x-2}\le0\)
\(\Leftrightarrow\frac{x^2-2x+1}{x-2}\le0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{x-2}\le0\Rightarrow\) không tồn tại x thỏa mãn
Vậy BPT có nghiệm duy nhất \(x=3\)
c/
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\4-x^2\ge0\\\sqrt{4-x^2}\ne1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\-2\le x\le2\\x\ne\pm\sqrt{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}1\le x\le2\\x\ne\sqrt{3}\end{matrix}\right.\)
BPT tương đương:
\(\frac{2\left(\sqrt{x-1}-2\right)}{\sqrt{4-x^2}-1}+\sqrt{x-1}-2\ge0\)
\(\Leftrightarrow\left(\sqrt{x-1}-2\right)\left(\frac{2}{\sqrt{4-x^2}-1}+1\right)\ge0\)
Do \(x\le2\Rightarrow\sqrt{x-1}\le1\Rightarrow\sqrt{x-1}-2< 0\)
BPt tương đương:
\(\frac{2}{\sqrt{4-x^2}-1}+1\le0\)
\(\Leftrightarrow\frac{1+\sqrt{4-x^2}}{\sqrt{4-x^2}-1}\le0\)
\(\Leftrightarrow\sqrt{4-x^2}-1< 0\) (do \(1+\sqrt{4-x^2}>0\) \(\forall x\))
\(\Leftrightarrow\sqrt{4-x^2}< 1\Leftrightarrow x^2>3\Rightarrow x>\sqrt{3}\)
Vậy nghiệm của BPT đã cho là: \(\sqrt{3}< x\le2\)
giải bpt
\(\left(\sqrt{x+4}-1\right)\sqrt{x+2}\ge\frac{x^3+4x^2+3x-2\left(x+3\right)\sqrt[3]{2x+3}}{\left(\sqrt[3]{2x+3}-3\right)\left(\sqrt{x+4}+1\right)}\)
giải bpt :
\(\frac{\sqrt{2\left(x^2+7x+3\right)}-\sqrt{x^2+x-6}-3\sqrt{x+1}}{1-2\sqrt{x^2-x+1}}\ge0\)
ĐKXĐ: \(x\ge2\)
Khi đó ta có \(x^2-x+1\ge3\Rightarrow1-2\sqrt{x^2-x+1}< 0\)
Do đó BPT tương đương:
\(\sqrt{2\left(x^2+7x+3\right)}-\sqrt{x^2+x-6}-3\sqrt{x+1}\le0\)
\(\Leftrightarrow\sqrt{2x^2+14x+6}\le\sqrt{x^2+x-6}+3\sqrt{x+1}\)
\(\Leftrightarrow2x^2+14x+6\le x^2+10x+3+6\sqrt{\left(x+1\right)\left(x^2+x-6\right)}\)
\(\Leftrightarrow x^2+4x+3\le6\sqrt{\left(x+1\right)\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\le6\sqrt{\left(x+1\right)\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+3\right)}\le6\sqrt{x-2}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\le36\left(x-2\right)\)
\(\Leftrightarrow x^2-32x+75\le0\)
\(\Rightarrow16-\sqrt{181}\le x\le16+\sqrt{181}\)
\(\sqrt{9x-27}+\sqrt{x-3}-\frac{1}{2}\sqrt{4x-12}\)
\(\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
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by
duc
giải pt
a) \(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(\sqrt{2x+3}-\sqrt{4-x}\right)^2-10\)
b) \(\sqrt{4x+1}+2\sqrt{1-x}+10\sqrt{-4x^2+3x+1}=13\)
c) \(\left(x^2+1\right)^2=13-x\sqrt{2x^2+4}\)
d) \(\left(\sqrt{x+1}+\sqrt{x-1}\right)^2-3=\frac{1}{\sqrt{x+1}-\sqrt{x-1}}\)
e) \(\left(\frac{2x-3}{\sqrt{x^2-1}}+2\right)\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x+1}}\right)=\frac{1}{x^2-1}\)
a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)
\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)
\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)
Đặt \(\sqrt{2x+3}+\sqrt{4-x}=a>0\Rightarrow a^2=x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\)
Phương trình trở thành:
\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)
\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)
Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(
b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)
Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)
\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}\)
Pt trở thành:
\(a+10\left(\frac{a^2-5}{4}\right)=13\)
\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)
\(\Leftrightarrow-4x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{4}\end{matrix}\right.\)
c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)
\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)
Đặt \(x\sqrt{2x^2+4}=a\) ta được:
\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)
d/ ĐKXĐ: \(x\ge1\)
Nhân cả tử và mẫu của vế phải với liên hợp của nó ta được:
\(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{x-1}\right)^2-3=\frac{\sqrt{x+1}+\sqrt{x+1}}{2}\)
Đặt \(\sqrt{x+1}+\sqrt{x-1}=a>0\)
\(\Rightarrow a^2-3=\frac{a}{2}\Rightarrow2a^2-a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}+\sqrt{x-1}=2\)
\(\Leftrightarrow x+\sqrt{x^2-1}=2\)
\(\Leftrightarrow\sqrt{x^2-1}=2-x\) (\(x\le2\))
\(\Leftrightarrow x^2-1=x^2-4x+4\)
\(\Rightarrow x=\frac{5}{4}\)
Giải bpt:
a,\(\frac{\sqrt{x^2-x+4}-2x-3}{x-2}>3\)
b, \(\sqrt{x\left(x-1\right)}+\sqrt{x\left(x+2\right)}\le\sqrt{x\left(4x+1\right)}\)
1.tính:\(\sqrt{35+\sqrt{69}}-\sqrt{35-\sqrt{69}}-\sqrt{12+8\sqrt{2}}\)
2.cho x,y,z\(\ne0\), xuz=100.tính
A\(=\)\(\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+10}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{10\sqrt{z}}{\sqrt{xz}+10\sqrt{z}+10}\)
3.giải pt : \(\sqrt{x-2\sqrt{x}+1}+\sqrt{x+2\sqrt{x}+1}=\frac{x+3}{2}\)
4.cho x,y>0 , \(xy=1\).CM: \(\frac{x^3}{1+y}+\frac{y^3}{1+x}\ge1\)
P/s: mình đag cần gấp ai giải đc cho 1 tick
1.
Xét riêng 2 căn lớn đầu tiên
Bình phương, thu gọn được căn(12-8 căn 2)
Giờ kết hợp kết quả này với căn lớn còn lại
Tiếp tục bình phương, thu gọn là xong
Mọi người giúp mình bài này với
Giải các bất phương trình sau (ưu tiên giải bằng phương pháp đặt ẩn phụ):
a, \(4 \sqrt{x}+\frac{2}{\sqrt{x}}<2 x+\frac{1}{2 x}+2\)
b, \(\frac{1}{1-x^{2}}>\frac{3 x}{\sqrt{1-x^{2}}}-1\)
c,\(\sqrt{\frac{1}{x^{2}}-\frac{3}{4}}<\frac{1}{x}-\frac{1}{2}\)
d, \(x+\frac{x}{\sqrt{x^{2}-1}}>\frac{35}{12}\)
Mình cảm ơn nhiều ạ.
a) \(4\sqrt{x}+\frac{2}{\sqrt{x}}< 2x+\frac{1}{2x}+2\)
hay \(2\sqrt{x}+\frac{1}{\sqrt{x}}< x+\frac{1}{4x}+1\)
\(\Leftrightarrow0< x+\frac{1}{4x}+1-2\sqrt{x}-\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow0< \left(\sqrt{x}\right)^2-2\sqrt{x}-2\sqrt{x}\cdot1+1+\frac{1}{\left(2\sqrt{x}\right)^2}-2\cdot\frac{1}{2\sqrt{x}}\)
\(\Leftrightarrow1< \left(\sqrt{x}-1\right)^2+\left(\frac{1}{2\sqrt{x}}-1\right)^2\)
\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}>1\\2\sqrt{x}>1\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>\frac{1}{4}\end{cases}\Rightarrow}x>1}\)
b) \(\frac{1}{1-x^2}>\frac{3}{\sqrt{1-x^2}}-1\left(1\right)\left(ĐK:-1< x< 1\right)\)
Ta có (1) <=> \(\frac{1}{1-x^2}-1-\frac{3x}{\sqrt{1-x^2}}+2>0\)\(\Leftrightarrow\frac{x^2}{1-x^2}-\frac{3x}{\sqrt{1-x^2}}+2>0\)
Đặt \(t=\frac{x}{\sqrt{1-x^2}}\)ta được
\(t^2-3t+2>0\Leftrightarrow\orbr{\begin{cases}\frac{x}{\sqrt{1-x^2}}< 1\\\frac{x}{\sqrt{1-x^2}}>2\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{1-x^2}>x\left(a\right)\\2\sqrt{1-x^2}< x\left(b\right)\end{cases}}}\)
(a) <=> \(\hept{\begin{cases}x< 0\\1-x^2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\1-x^2>x^2\end{cases}}}\)
\(\Leftrightarrow-1< x< 0\)hoặc \(\hept{\begin{cases}x\ge0\\x^2< \frac{1}{2}\end{cases}}\)
\(\Leftrightarrow-1< x< 0\)hoặc \(0\le x\le\frac{\sqrt{2}}{2}\Leftrightarrow-1< x< \frac{\sqrt{2}}{2}\)
(b) \(\Leftrightarrow\hept{\begin{cases}1-x^2>0\\x>0\\4\left(1-x^2\right)< x^2\end{cases}\Leftrightarrow\hept{\begin{cases}0< x< 1\\x^2>\frac{4}{5}\end{cases}\Leftrightarrow}\frac{2}{\sqrt{5}}< x< 1}\)
ok đợi nấu ăn xong r làm cho
a) điều kiện x>0
khi đó
\(\left(a\right)\Leftrightarrow4\left(\sqrt{4}+\frac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)^2\)
\(\Leftrightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}>2\Leftrightarrow2x-4\sqrt{x}+1>0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}< \frac{2-\sqrt{2}}{2}\\\sqrt{x}>\frac{2+\sqrt{2}}{2}\end{cases}}\)