Giải phương trình: x(x+5)=\(2\sqrt[3]{\left(x-1\right)\left(x+6\right)}-2\)
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9 tháng 11 2023
Bài 1:
\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=4-\sqrt{5}+\sqrt{5}+1=5\)
Bài 2:
a: ĐKXĐ: x>=3
\(\sqrt{x-3}=6\)
=>x-3=36
=>x=36+3=39(nhận)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(x-3\right)^2}=12\)
=>\(\left|x-3\right|=12\)
=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Bài 3:
a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)
\(=\dfrac{3-x\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\cdot\dfrac{3-\sqrt{x}}{3-x}\)
\(=\dfrac{3-x\sqrt{x}+3\sqrt{x}-x}{3-x}\)
\(=\dfrac{-\sqrt{x}\left(x-3\right)-\left(x-3\right)}{-\left(x-3\right)}=\dfrac{\left(x-3\right)\left(\sqrt{x}+1\right)}{x-3}=\sqrt{x}+1\)
b: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)
\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)
\(=\sqrt{3x-1}+\sqrt{5}\)
d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\left(a-2\right)}{a+2}\)
HP
6 tháng 1 2021
ĐK: \(x\ge1\)
\(pt\Leftrightarrow2\sqrt{\left(x-1\right)\left(x+2\right)}-\sqrt{x-1}-6\sqrt{x+2}+3=0\)
\(\Leftrightarrow\left(2\sqrt{x+2}-1\right)\left(\sqrt{x-1}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+2}=1\\\sqrt{x-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x+2\right)=1\\x-1=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{4}\left(l\right)\\x=10\left(tm\right)\end{matrix}\right.\)
Vậy ...
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 1 2021
Xét \(f\left(x;y;z\right)=\left(3x+4y+5z\right)^2-44\left(xy+yz+zx\right)\)
\(=\left(y+2z+3\right)^2-44yz-44\left(y+z\right)\left(1-y-z\right)\)
\(=45y^2+2y\left(24z-19\right)+48z^2-32z+9\)
\(\Delta_y'=\left(24z-9\right)^2-45\left(48z^2-32z+9\right)=-44\left(6z-1\right)^2\le0\)
\(\Rightarrow f\left(x;y;z\right)\ge0\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 2 2021
Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:
\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)
\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)
Ta có:
\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)
\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)
Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)
Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)
1 tháng 10 2021
Tham khảo:
1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24
A
1
XO
11 tháng 4 2023
ĐKXĐ : \(x\ge-2\)
\(\sqrt{1+\left(x+2\right).\sqrt{1+\left(x+3\right).\left(x+5\right)}}=2023x+1\)
\(\Leftrightarrow\sqrt{1+\left(x+2\right).\sqrt{x^2+8x+16}}=2023x+1\)
\(\Leftrightarrow\sqrt{1+\left(x+2\right).\left(x+4\right)}=2023x+1\) (Do \(x\ge-2\Rightarrow x+4>0\))
\(\Leftrightarrow\sqrt{x^2+6x+9}=2023x+1\)
\(\Leftrightarrow x+3=2023x+1\) (Do \(x\ge-2\Rightarrow x+3>0\)
\(\Leftrightarrow x=\dfrac{1}{1011}\)(tm)
Vậy tập nghiệm \(S=\left\{\dfrac{1}{1011}\right\}\)
X = - 1,056501731
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