\(\sqrt{x+15}+\sqrt{x+8}=4\sqrt[3]{x}-3\)
tìm MIN của \(\sqrt{x+2\sqrt{x-1}}+3\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+15-8\sqrt{x-1}}\)
3) tìm x biết
a) \(\sqrt{x+9}=7\)
b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\)
c) \(\sqrt{x^2-6x+9}=2x+1\)
d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\)
lm nhanh giúp mk nhé mk đang cần gấp
Lời giải:
a. ĐKXĐ: $x\geq -9$
PT $\Leftrightarrow x+9=7^2=49$
$\Leftrightarrow x=40$ (tm)
b. ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$
$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$
$\Leftrgihtarrow 3\sqrt{2x+3}=15$
$\Leftrightarrow \sqrt{2x+3}=5$
$\Leftrightarrow 2x+3=25$
$\Leftrightarrow x=11$ (tm)
c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{2}{3}\)
d. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)
\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)
\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)
\(\Leftrightarrow -1=9\) (vô lý)
Vậy pt vô nghiệm.
a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)
b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)
\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)
\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)
\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)
c) \(\sqrt{x^2-6x+9}=2x+1\)
Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)
\(\Rightarrow\sqrt{\left(x-3\right)^2}=2x+1\Rightarrow\left|x-3\right|=2x+1\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)
d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)
\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)
\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)
\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)
\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)
\(\dfrac{8\sqrt{x}-x-31}{x-8\sqrt{x}-15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\) rút gọn giùm ae
Sửa đề: \(\dfrac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)
Ta có: \(\dfrac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)
\(=\dfrac{-x+8\sqrt{x}-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}-\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}+\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{-x+8\sqrt{x}-31-\left(x-25\right)+3x-9\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{2x-2\sqrt{x}-28-x+25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x-3\sqrt{x}+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)
1)\(\sqrt{4+2x-x^2}=x-2\)
2)\(\sqrt{25-x^2}=x-1\)
3)(x+4).\(\sqrt{10-x^2}=x^2+2x-8\)
4)(x-3).\(\sqrt{x^2-3x+2}=x^2-8x+15\)
5)\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x-6\sqrt{x-1}+8}=1\)
6)\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)
7)\(^{x^2+x-2\sqrt{x+1}+2=0}\)
8)x-4\(\sqrt{2x+4}-2\sqrt{1-x}+10=0\)
1.
ĐK: $-x^2+2x+4\geq 0$
PT \(\Rightarrow \left\{\begin{matrix} x-2\geq 0\\ 4+2x-x^2=(x-2)^2=x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 2\\ 6x=2x^2\end{matrix}\right.\Rightarrow x=3\) (thỏa mãn)
Vậy...........
2)
ĐK: $-5\leq x\leq 5$
PT \(\Rightarrow \left\{\begin{matrix} x-1\geq 0\\ 25-x^2=(x-1)^2=x^2-2x+1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 2x^2-2x-24=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x^2-x-12=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ (x+3)(x-4)=0\end{matrix}\right.\)
\(\Rightarrow x=4\) (thỏa mãn)
3)
ĐK: $x^2\leq 10$
PT $\Leftrightarrow (x+4)\sqrt{10-x^2}=(x+4)(x-2)$
$\Leftrightarrow (x+4)[\sqrt{10-x^2}-(x-2)]=0$
Nếu $x+4=0\Rightarrow x=-4$ (không thỏa mãn ĐKXĐ)
Nếu $\sqrt{10-x^2}-(x-2)=0$
$\Leftrightarrow \sqrt{10-x^2}=x-2$
\(\Rightarrow \left\{\begin{matrix} x-2\geq 0\\ 10-x^2=(x-2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 2\\ 2x^2-4x-6=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ 2(x-3)(x+1)=0\end{matrix}\right.\Rightarrow x=3\)
\(\sqrt{x+3-4\sqrt{x-1}}\)+ \(\sqrt{x+15-8\sqrt{x-1}}\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+15-8\sqrt{x-1}}\)
\(=\sqrt{\left(x-1\right)-4\sqrt{x-1}+4}+\sqrt{\left(x-1\right)-8\sqrt{x-1}+16}\)
\(=\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-4\right)^2}\)
\(=\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-4\right|\)
Bài 1 : Rút gọn biểu thức sau :
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
Bài 2 : Chứng minh đẳng thức sau :
\(\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}=2\sqrt{5}-2\)
Bài 3 : Cho biểu thức E = \(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\)
a) Rút gọn biẻu thức E
b) Tính giá trị của E khi x = \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
1,\(\sqrt{x-5}+\sqrt{x+4}=3\)
2,\(\sqrt{x+4-4\sqrt{x}}+\sqrt{x+6-6\sqrt{x}}=1\)
3,\(\sqrt{x+3}-\sqrt{x-4}=1\)
4,\(\sqrt{15-x}+\sqrt{3-x}=6\)
5,\(\sqrt{10-x}+\sqrt{x+3}=5\)
6,\(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\)
7,\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\)
8,\(\sqrt{x^2-5x+6}+\sqrt{x-2-3\sqrt{x-3}}=3\)
9,\(2x^2-x+4=2\sqrt{2x+3}\)
mầy câu 1;3;;4;5 cách làm nhu nhau(nhân liên hop hoac bình phuong lên)
1.
\(DK:x\in\left[-4;5\right]\)
\(\Leftrightarrow\sqrt{x-5}+\left(\sqrt{x+4}-3\right)=0\)
\(\Leftrightarrow\sqrt{x-5}+\frac{x-5}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\sqrt{x-5}\left(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}\right)=0\)
Vi \(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}>0\)
\(\Rightarrow\sqrt{x-5}=0\)
\(x=5\left(n\right)\)
Vay nghiem cua PT la \(x=5\)
2.
\(DK:x\ge0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow|\sqrt{x}-2|+|\sqrt{x}-3|=1\)
Ta co:
\(|\sqrt{x}-2|+|\sqrt{x}-3|=|\sqrt{x}-2|+|3-\sqrt{x}|\ge|\sqrt{x}-2+3-\sqrt{x}|=1\)
Dau '=' xay ra khi \(\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x}-2\ge0\\3-\sqrt{x}\ge0\end{cases}\Leftrightarrow4\le x\le9\left(n\right)}\)
TH2:(loai)
Vay nghiem cua PT la \(x\in\left[4;9\right]\)
6.
\(DK:x\ge2\)
\(\Leftrightarrow\left(\sqrt{2x-1}-\sqrt{x+1}\right)+\sqrt{x-2}=0\)
\(\Leftrightarrow\frac{x-2}{\sqrt{2x-1}+\sqrt{x+1}}+\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\frac{\sqrt{x-2}}{\sqrt{2x-1}+\sqrt{x+1}}+1\right)=0\)
Vi \(\frac{\sqrt{x-2}}{\sqrt{2x-1}+\sqrt{x+1}}+1>0\)
\(\Rightarrow x=2\left(n\right)\)
Vay nghiem cua PT la \(x=2\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
Giải phương trình vô tỉ:
1/ \(\sqrt{x^2+12}+5=3x+\sqrt{x^2+15}\)
2/ \(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x+1\right)}-\sqrt{x^2-3x+4}\)
3/ \(\sqrt[5]{x-1}+\sqrt[3]{x+8}=-x^3+1\)
4/ \(\sqrt{5-x^6}+\sqrt[3]{3x^4-2}=1\)