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\(VT=\sqrt{\left(3x+1\right)^2+\left(2x-3\right)^2}+\sqrt{\left(2x-\frac{5}{2}\right)^2+\left(x-\frac{3}{2}\right)^2}+\sqrt{x^2+\left(4x-6\right)^2}\)
\(\ge\sqrt{\left(3x+1\right)^2}+\sqrt{\left(2x-\frac{5}{2}\right)^2}+\sqrt{x^2}=\left|3x+1\right|+\left|2x-\frac{5}{2}\right|+\left|x\right|\)
\(\ge\left|3x+1+2x-\frac{5}{2}+x\right|=\left|6x-\frac{3}{2}\right|\ge6x-\frac{3}{2}\)
Dấu "=" xảy ra khi x = \(\frac{3}{2}\)
\(VP=\frac{1}{2}\left[-2\left(2x-3\right)^2+12x-3\right]\le\frac{1}{2}\left(12x-3\right)=6x-\frac{3}{2}\)
Dấu "=" xảy ra khi x = \(\frac{3}{2}\)
Từ đó suy ra nghiệm phương trình là \(x=\frac{3}{2}\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
Ta có:
\(\sqrt{x^2-x+19}+\sqrt{7x^2+8x+13}+\sqrt{13x^2+17x+7}\)
\(=\sqrt{\frac{1}{4}\left(2x-1\right)^2+\frac{75}{4}}+\sqrt{\left(2x-1\right)^2+3\left(x+2\right)^2}+\sqrt{\frac{1}{4}\left(2x-1\right)^2+\frac{3}{4}\left(4x+3\right)^2}\)
\(\ge\sqrt{\frac{75}{4}}+\sqrt{3\left(x+2\right)^2}+\sqrt{\frac{3}{4}\left(4x+3\right)^2}\)
\(=\frac{5\sqrt{3}}{2}+\sqrt{3}\left(x+2\right)+\frac{\sqrt{3}\left(4x+3\right)}{2}=3\sqrt{3}\left(x+2\right)\)
Dấu = xảy ra khi ....
Ta viết lại phương trình thành:
\(\left(2x-1\right)^3-\left(x^2-x-1\right)=\left(x+1\right)\sqrt[3]{\left(x+1\right)\left(2x-1\right)+x^2-x-1}\)
Đặt: \(a=2x-1;b=\sqrt[3]{\left(x+1\right)\left(2x-1\right)+x^2-x-1}=\sqrt[3]{3x^2-2}\) ta thu được hệ phương trình:
\(\hept{\begin{cases}a^3-\left(x^2-x+1\right)=\left(x+1\right)b\\b^3-\left(x^2-x+1\right)=\left(x+1\right)a\end{cases}}\)
Trừ 2 pt của hệ cho nhau ta được: \(\left(a-b\right)\left(a^2+ab+b^2+x+1\right)=0\)
Trường hợp 1: \(a=b\) ta có:
\(2x-1=\sqrt[3]{3x^2-2}\Leftrightarrow8x^3-15x^2+6x+1=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{8}\end{cases}}\)
Trường hợp 2: \(a^2+ab+b^2+x+1=0\Leftrightarrow\left(a+\frac{b}{2}\right)^2+\frac{3}{4}\left(2x-1\right)^2+x+1=0\)
\(\Leftrightarrow4\left(a+\frac{b}{2}\right)^2+4x^2+2\left(2x-1\right)^2+5=0\left(vn\right)\)
Vậy pt có 2 nghiệm là: \(x=1;x=-\frac{1}{8}\)
a. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$
$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$
$\Leftrightarrow -\sqrt{x-1}=-17$
$\Leftrightarrow \sqrt{x-1}=17$
$\Leftrightarrow x-1=289$
$\Leftrightarrow x=290$
b. ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$
$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$
$\Leftrihgtarrow \sqrt{2x-1}=2$
$\Leftrightarrow x=2,5$ (tm)
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm
a,\(\sqrt{\left(3x-1\right)^2}=5=>|3x-1|=5=>\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b, \(\sqrt{4x^2-4x+1}=3=\sqrt{\left(2x-1\right)^2}=3=>\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c, \(\sqrt{x^2-6x+9}+3x=4=>|x-3|=4-3x\)
TH1: \(|x-3|=x-3< =>x\ge3=>x-3=4-3x=>x=1,75\left(ktm\right)\)
TH2 \(|x-3|=3-x< =>x< 3=>3-x=4-3x=>x=0,5\left(tm\right)\)
Vậy x=0,5...
d, đk \(x\ge-1\)
=>pt đã cho \(< =>9\sqrt{x+1}-6\sqrt{x+1}+4\sqrt{x+1}=12\)
\(=>7\sqrt{x+1}=12=>x+1=\dfrac{144}{49}=>x=\dfrac{95}{49}\left(tm\right)\)
a) Ta có: \(\sqrt{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow\left|3x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b) Ta có: \(\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c) Ta có: \(\sqrt{x^2-6x+9}+3x=4\)
\(\Leftrightarrow\left|x-3\right|=4-3x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-23x\left(x\ge3\right)\\x-3=23x-4\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+23x=4+3\\x-23x=4+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{24}\left(loại\right)\\x=\dfrac{-4}{22}=\dfrac{-2}{11}\left(loại\right)\end{matrix}\right.\)
