a)\(\sqrt{2x-5}\)+ 2\(\sqrt{7-x}\)=\(\sqrt{3}\).x2+\(\sqrt{3}.8x+19\sqrt{3}\)
b)\(\sqrt{4x+x^2}+\sqrt{4x-x^2}=4x+1\)
Mọi người giúp mình với
Bài 1 GIẢI PHƯƠNG TRÌNH:
a) \(\sqrt{x-5}=\sqrt{3-x}\)
b) \(\sqrt{4-5x}=\sqrt{2-5x}\)
c) x2+4x+5=2\(\sqrt{2x+3}\)
d) \(\sqrt{x^2-2x+1}=\sqrt{4x^2-4x+1}\)
\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy pt vô nghiệm
\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)
\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
a) \(\sqrt{x-5}=\sqrt{3-x}\)
⇔\(\left(\sqrt{x-5}\right)^2=\left(\sqrt{3-x}\right)^2\)
⇔\(x-5=3-x\)
⇔\(x=4\)
b) \(\sqrt{4-5x}=\sqrt{2-5x}\)
⇔\(\left(\sqrt{4-5x}\right)^2=\left(\sqrt{2-5x}\right)^2\)
⇔\(4-5x=2-5x\)
⇔\(2=0\) (Vô lí)
1/ \(\lim\limits_{x\to 1}\) \(\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}\)
2/ \(\lim\limits_{x \to \ +\infty} \)\(x\left[\sqrt{4x^2+5}-\sqrt[3]{8x^3-1}\right]\)
3/ \(\lim\limits_{x\to 1}\)\(\dfrac{x^3-2x-1}{x^5-2x-1}\)
Giải giúp mình với ạ
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{7+x^3}-2\right)-\left(\sqrt{3+x^2}-2\right)}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^3-1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x^2-1}{\sqrt{3+x^2}+2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^2+x+1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x+1}{\sqrt{3+x^2}+2}}{1}=\dfrac{3}{12}-\dfrac{2}{4}=\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\).
giải phương trình
a)\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
b)\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
c)\(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\)
d)\(\dfrac{1}{3}\sqrt{2x}-\sqrt{8x}+\sqrt{18x}-10=2\)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
giải pt
a>\(\sqrt{2x+3}+\sqrt{5-8x}=\sqrt{4x+7}\)
b>\(\sqrt[3]{x-17}+\sqrt{x+8}=5\)
c>\(\sqrt[3]{4x-y^2}-\sqrt[3]{2+x}=\sqrt[3]{4x^2}\)
\(b.\sqrt[3]{x-17}+\sqrt{x+8}=5\) \(\left(ĐK:x\ge-8\right)\)
Đặt: \(a=\sqrt[3]{x-17},b=\sqrt{x+8}\)
\(\Rightarrow x-17=a^3,x+8=b^2\)
Khi đó:
\(\left\{{}\begin{matrix}a+b=5\\a^3-b^2=x-17-x-8=-25\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\a^3-b^2=-25\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\\left(5-b\right)^3-b^2=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\b^3-14b^2+75b-150=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\b^3-5b^2-9b^2+45b+30b-150=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\b^2\left(b-5\right)-9b\left(b-5\right)+30\left(b-5\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\\left(b-5\right)\left(b^2-9b+30\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\\left[{}\begin{matrix}b=5\\b^2-9b+30=\left(b-\dfrac{9}{2}\right)^2+\dfrac{39}{4}=0\left(l\right)\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=5\end{matrix}\right.\)
Thế vào ta được:
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt[3]{x-17}=0\\\sqrt{x+8}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-17=0\\x+8=25\end{matrix}\right.\) \(\Leftrightarrow x=17\left(n\right)\)
Câu (a)
Huhu hack não quá đuy :v không biết làm đúng or sai nữa :v dù sao cũng là 20p của mừn đó
Giải phương trình:(Nhớ tìm điều kiện)
a) \(\sqrt{2x-1}=\sqrt{5}\)
b)\(\sqrt{x-5}\) = 3
c)\(\sqrt{4x^2+4x+1}=6\)
d)\(\sqrt{\left(x-3\right)^2}=3-x\)
e)\(\sqrt{2x+5}=\sqrt{1-x}\)
f)\(\sqrt{x^2-x}=\sqrt{3-x}\)
g)\(\sqrt{2x^2-3}=\sqrt{4x-3}\)
h)\(\sqrt{2x-5}=\sqrt{x-3}\)
i)\(\sqrt{x^2-x+6}=\sqrt{x^2+3}\)
a, ĐKXĐ : \(x\ge\dfrac{1}{2}\)
PT <=> 2x - 1 = 5
<=> x = 3 ( TM )
Vậy ...
b, ĐKXĐ : \(x\ge5\)
PT <=> x - 5 = 9
<=> x = 14 ( TM )
Vậy ...
c, PT <=> \(\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy ...
d, PT<=> \(\left|x-3\right|=3-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=x-3\\x-3=3-x\end{matrix}\right.\)
Vậy phương trình có vô số nghiệm với mọi x \(x\le3\)
e, ĐKXĐ : \(-\dfrac{5}{2}\le x\le1\)
PT <=> 2x + 5 = 1 - x
<=> 3x = -4
<=> \(x=-\dfrac{4}{3}\left(TM\right)\)
Vậy ...
f ĐKXĐ : \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)
PT <=> \(x^2-x=3-x\)
\(\Leftrightarrow x=\pm\sqrt{3}\) ( TM )
Vậy ...
a) \(\sqrt{2x-1}=\sqrt{5}\) (x \(\ge\dfrac{1}{2}\))
<=> 2x - 1 = 5
<=> x = 3 (tmđk)
Vậy S = \(\left\{3\right\}\)
b) \(\sqrt{x-5}=3\) (x\(\ge5\))
<=> x - 5 = 9
<=> x = 4 (ko tmđk)
Vậy x \(\in\varnothing\)
c) \(\sqrt{4x^2+4x+1}=6\) (x \(\in R\))
<=> \(\sqrt{\left(2x+1\right)^2}=6\)
<=> |2x + 1| = 6
<=> \(\left[{}\begin{matrix}\text{2x + 1=6}\\\text{2x + 1}=-6\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)(tmđk)
Vậy S = \(\left\{\dfrac{5}{2};\dfrac{-7}{2}\right\}\)
Giai phuong trinh
1/ \(\sqrt{x^2+4x+5}+\sqrt{x^2-6x+13}=3\)
2/ \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=6x-x^2-5\)
3/ \(\sqrt{2x^2-4x+27}+\sqrt{3x^2-6x+12}=4x^2+8x+4\)
4/ \(\sqrt{x^2+x+7}+\sqrt{x^2+x+2}=\sqrt{3x^2+3x+19}\)
5/ \(\left(x+2\right)\left(x+3\right)-\sqrt{x^2+5x+1}=9\)
6/ \(\left(x+4\right)\left(x+1\right)-3\sqrt{x^2+5x+2}=6\)
7/ \(\sqrt{2x^2+3x+5}+\sqrt{2x^2-3x+5}=3\sqrt{x}\)
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
Giải các phương trình sau: (hệ phương trình)
1.\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)
2.\(\sqrt{3-4x}+\sqrt{4x+1}=-16x^2-8x+1\)
3. \(\sqrt{x^2-2x+5}+\sqrt{x+1}=2\)
4. \(\left(-4x-1\right)\sqrt{x^2+1}=2x^2+2x+1\)
5. \(\sqrt{-4x-1}+\sqrt{4x^2+8x+3}=-4x^2-4x\)
6. \(\left(x-3\right)\left(x+1\right)+4\left(x-3\right)\sqrt{\frac{x+1}{x-3}}=-3\)
7. \(\sqrt{x\left(x-1\right)}+\sqrt{x\left(x+2\right)}=2\sqrt{x^2}\)
Ai làm được 4 bài hoặc nhiều hơn mik sẽ tick nha :)
a) 4sqrt(2x + 1) - sqrt(8x + 4) + 1/2 * sqrt(32x + 16) = 12 b) sqrt(4x ^ 2 - 4x + 1) = 5 . c) (2sqrt(x) - 3)/(sqrt(x) - 1) = - 1/2
a) \(4\sqrt{2x+1}-\sqrt{8x+4}+\dfrac{1}{2}\sqrt{32x+16}=12\) (ĐK: \(x\ge-\dfrac{1}{2}\))
\(\Leftrightarrow4\sqrt{2x+1}-\sqrt{4\left(2x+1\right)}+\dfrac{1}{2}\cdot4\sqrt{2x+1}=12\)
\(\Leftrightarrow4\sqrt{2x+1}-2\sqrt{2x+1}+2\sqrt{2x+1}=12\)
\(\Leftrightarrow4\sqrt{2x+1}=12\)
\(\Leftrightarrow\sqrt{2x+1}=\dfrac{12}{4}\)
\(\Leftrightarrow2x+1=3^2\)
\(\Leftrightarrow2x=9-1\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=\dfrac{8}{2}\)
\(\Leftrightarrow x=4\left(tm\right)\)
b) \(\sqrt{4x^2-4x+1}=5\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\)
\(\Leftrightarrow\left|2x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\left(x\ge\dfrac{1}{2}\right)\\2x-1=-5\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{2}\\x=-\dfrac{4}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
c) \(\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}=-\dfrac{1}{2}\)(ĐK: \(x\ge0;x\ne1\))
\(\Leftrightarrow-\left(\sqrt{x}-1\right)=2\left(2\sqrt{x}-3\right)\)
\(\Leftrightarrow-\sqrt{x}+1=4\sqrt{x}-6\)
\(\Leftrightarrow4\sqrt{x}+\sqrt{x}=1+6\)
\(\Leftrightarrow5\sqrt{x}=7\)
\(\Leftrightarrow\sqrt{x}=\dfrac{7}{5}\)
\(\Leftrightarrow x=\dfrac{49}{25}\left(tm\right)\)
giải phương trình
a) 1+\(\sqrt{x^2+7x+10}\)=\(\sqrt{x+5}\)+\(\sqrt{x+2}\)
b) \(\sqrt{4x^2-2x+\dfrac{1}{4}}\)=\(4x^3\)-\(x^2\)+8x-2
a:
ĐKXĐ: \(x>=-2\)
\(1+\sqrt{x^2+7x+10}=\sqrt{x+5}+\sqrt{x+2}\)
=>\(1+\sqrt{\left(x+2\right)\left(x+5\right)}=\sqrt{x+5}+\sqrt{x+2}\)
Đặt \(\sqrt{x+5}=a;\sqrt{x+2}=b\)(ĐK: a>0 và b>0)
Phương trình sẽ trở thành:
1+ab=a+b
=>ab-a-b+1=0
=>a(b-1)-(b-1)=0
=>(b-1)(a-1)=0
=>\(\left\{{}\begin{matrix}a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow a=b=1\)
=>\(\left\{{}\begin{matrix}x+5=1\\x+2=1\end{matrix}\right.\)
=>\(x\in\varnothing\)
b: \(\sqrt{4x^2-2x+\dfrac{1}{4}}=4x^3-x^2+8x-2\)
=>\(\sqrt{\left(2x\right)^2-2\cdot2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)
=>\(\sqrt{\left(2x-\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)
=>\(\left|2x-\dfrac{1}{2}\right|=4x^3-x^2+8x-2\)(1)
TH1: x>=1/4
\(\left(1\right)\Leftrightarrow4x^3-x^2+8x-2=2x-\dfrac{1}{2}\)
=>\(4x^3-x^2+6x-\dfrac{3}{2}=0\)
=>\(x^2\left(4x-1\right)+1,5\left(4x-1\right)=0\)
=>\(\left(4x-1\right)\left(x^2+1,5\right)=0\)
=>4x-1=0
=>x=1/4(nhận)
TH2: x<1/4
Phương trình (1) sẽ trở thành:
\(4x^3-x^2+8x-2=-2x+\dfrac{1}{2}\)
=>\(x^2\left(4x-1\right)+2\left(4x-1\right)+0,5\left(4x-1\right)=0\)
=>\(\left(4x-1\right)\cdot\left(x^2+2,5\right)=0\)
=>4x-1=0
=>x=1/4(loại)