4.giai phuong trinh:
a.\(\sqrt{2}.x-\sqrt{6}=0\)
b.\(\frac{x^2}{\sqrt{3}}-\sqrt{12}=0\)
c.\(\sqrt{3.x}+\sqrt{3}=\sqrt{12}+\sqrt{27}\)
Giải phương trình
a \(\sqrt{x^2-4}-3\sqrt{x-2}=0\)
b \(x-6\sqrt{x}+9=0\)
c \(\sqrt{9x-27}+\sqrt{x-3}-\frac{1}{2}\sqrt{4x-12}=7\)
d \(3\sqrt{8x+4}-\frac{1}{3}\sqrt{18x+9}-\frac{1}{2}\sqrt{50x+25}+\sqrt[]{\frac{2x+1}{4}}=6\)
Lời giải:
a) ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{(x-2)(x+2)}-3\sqrt{x-2}=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x+2}-3)=0$
\(\Rightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x+2}-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=7\end{matrix}\right.\) (thỏa mãn)
Vậy..........
b) ĐK: $x\geq 0$
PT $\Leftrightarrow (\sqrt{x}-3)^2=0$
$\Leftrightarrow \sqrt{x}-3=0$
$\Leftrightarrow x=9$ (thỏa mãn)
c) ĐK: $x\geq 3$
PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4(x-3)}=7$
$\Leftrightarrow 3\sqrt{x-3}+\sqrt{x-3}-\sqrt{x-3}=7$
$\Leftrightarrow 3\sqrt{x-3}=7$
$\Leftrightarrow x-3=(\frac{7}{3})^2$
$\Rightarrow x=\frac{76}{9}$
d)
ĐK: $x\geq \frac{-1}{2}$
PT $\Leftrightarrow 3\sqrt{4(2x+1)}-\frac{1}{3}\sqrt{9(2x+1)}-\frac{1}{2}\sqrt{25(2x+1)}+\sqrt{\frac{1}{4}(2x+1)}=6$
$\Leftrightarrow 6\sqrt{2x+1}-\sqrt{2x+1}-\frac{5}{2}\sqrt{2x+1}+\frac{1}{2}\sqrt{2x+1}=6$
$\Leftrightarrow 3\sqrt{2x+1}=6$
$\Leftrightarrow \sqrt{2x+1}=2$
$\Rightarrow x=\frac{3}{2}$ (thỏa mãn)
a.\(\sqrt{2}.x-\sqrt{50}=0\)
b.\(\sqrt{3}.x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)
c.\(\sqrt{3}.x^2-\sqrt{12}=0\)
d.\(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\)
a) \(\sqrt{2}\cdot x-\sqrt{50}=0< =>\sqrt{2}\cdot x=\sqrt{50}\)
<=> x= 5
b) \(\sqrt{3}\cdot x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)
<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot\sqrt{4}+\sqrt{3}\cdot\sqrt{9}\)
<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot5< =>x+1=5\)
<=> x=4
c) \(\sqrt{3}\cdot x^2-\sqrt{12}=0\\ < =>x^2=\sqrt{4}=2;-2\\ < =>x=\sqrt{2};-\sqrt{2}\)
d) \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\\ < =>x^2=\sqrt{100}=10;-10\\ < =>x=\sqrt{10};-\sqrt{10}\)
bài 1: thực hiện phép tính
a, (\(\sqrt{12}+3\sqrt{15}-4\sqrt{135}\)).\(\sqrt{3}\)
b, A=\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
c, \(\frac{9\sqrt{5^2+3\sqrt{27}}}{\sqrt{5}+\sqrt{3}}\)
d, \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
e, (\(\sqrt{12}+\sqrt{15}+\sqrt{27}\)):\(\sqrt{15}\)
f, (12\(\sqrt{50}-8\sqrt{200}+7\sqrt{450}\)):\(\sqrt{10}\)
g, (\(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{\frac{9}{7}}\)):\(\sqrt{7}\)
bài 2:rút gọn rồi tính các giá trị biểu thức
a, A= \(\sqrt{\frac{\left(x-6\right)^4}{\left(5-x\right)^2}}\)+\(\frac{x^2-36}{x-5}\) (x<5) tại x=4
b, B=5x-\(\sqrt{125}\)+\(\frac{\sqrt{x^3+5x^2}}{\sqrt{x+5}}\) (x ≥ 0)tại x=\(\sqrt{5}\)
giải pt
a) \(x\sqrt{x^2-4x+3}=x^2+x\)
b) \(x^2+x-12-\left(x-3\right)\sqrt{10-x^2}=0\)
c) \(\sqrt{6+x-x^2}=\frac{\left(2x+5\right)\sqrt{6+x-x^2}}{x+4}\)
d) \(\sqrt{\frac{12+x-x^2}{2x+9}}-\frac{\sqrt{12+x-x^2}}{x+3}=0\)
e) \(\sqrt{x^3}+\sqrt{x^3+x^2+2x}=3\sqrt{x}\)
a, ĐK:\(x^2-4x+3\ge0\Rightarrow\left[{}\begin{matrix}x\le1\\3\le x\end{matrix}\right.\)
\(PT\Leftrightarrow x\sqrt{x^2-4x+3}=x\left(x+1\right)\)
Với x = 0 \(\Rightarrow pttm\)
Với \(x\ne0\) \(\Rightarrow\sqrt{x^2-4x+3}=x+1\)
\(\Rightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-4x+3=x^2+2x+1\end{matrix}\right.\)\(\Rightarrow x=\frac{1}{3}\left(tm\right)\)
b,ĐK: \(-\sqrt{10}\le x\le\sqrt{10}\)
\(PT\Leftrightarrow\left(x-3\right)\left(x+4\right)-\left(x-3\right)\sqrt{10-x^2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x+4-\sqrt{10-x^2}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x+4=\sqrt{10-x^2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+8x+16=10-x^2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2+4x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\end{matrix}\right.\)(tm)
c/ ĐKXĐ: \(-2\le x\le3\)
\(\Leftrightarrow\left(x+4\right)\sqrt{6+x-x^2}-\left(2x+5\right)\sqrt{6+x-x^2}=0\)
\(\Leftrightarrow\sqrt{6+x-x^2}\left(x+4-2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+6=0\\-x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-1\\x=3\end{matrix}\right.\)
d/ ĐKXĐ: \(3< x\le4\)
\(\Leftrightarrow\sqrt{-x^2+x+12}\left(\frac{1}{\sqrt{2x+9}}-\frac{1}{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+12=0\\\sqrt{2x+9}=x+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+12=0\\2x+9=x^2+6x+9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+12=0\\x^2+4x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\x=4\\x=0\\x=-4\left(l\right)\end{matrix}\right.\)
e/ ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow\sqrt{x}\left(x+\sqrt{x^2+x+2}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\Rightarrow x=0\\\sqrt{x^2+x+2}=3-x\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\x^2+x+2=\left(3-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\7x=7\end{matrix}\right.\) \(\Rightarrow x=1\)
bài 1 thực hiện các phép tính sau:
a) \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
b)\(2\sqrt{2}-\sqrt{3^2}\)
c) \(\left(\sqrt{x}-3\right)\left(4-\sqrt{x}\right)\)
bài 2 rút gọn biểu thức sau:
a)\(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
b)\(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
c)\(\frac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}\)
d)\(\frac{x-2\sqrt{x}}{x+4-4\sqrt{x}}\)
giai cac phuong trinh
a)\(2x^4+5x^3+x^2+5x+2=0\)
b)\(\sqrt{x-1}-\sqrt[3]{2-x}=1\)
c)\(x-\sqrt{x}+1=\sqrt{2x^2-30x+2}\)
d)\(2x^2+3x+7=\left(x-5\right)\sqrt{2x^2+1}\)
e)\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)
Giải PT:
a) -5x+7\(\sqrt{x}\) +12=0
b) \(\dfrac{1}{3}\)\(\sqrt{4x^2-20}\) +2\(\sqrt{\dfrac{x^2-5}{9}}\) -3\(\sqrt{x^2-5}=0\)
c) \(\sqrt{9x+27}+5\sqrt{x+3}-\dfrac{3}{4}\sqrt{16x+48}=5\)
d) \(\sqrt{49x-98}-14\sqrt{\dfrac{x-2}{49}}=3\sqrt{x-2}+8\)
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$
$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$
$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$
Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$
$\Leftrightarrow \sqrt{x}=\frac{12}{5}$
$\Leftrightarrow x=5,76$ (thỏa mãn)
d. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$
$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$
$\Leftrightarrow 2\sqrt{x-2}=8$
$\Leftrightarrow \sqrt{x-2}=4$
$\Leftrightarrow x=4^2+2=18$ (tm)
b. ĐKXĐ: $x^2\geq 5$
PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$
$\Leftrightarrow \sqrt{x^2-5}=0$
$\Leftrightarrow x=\pm \sqrt{5}$
Giải phương trình và bất phương trình:
a) \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}-3=0}\)
b) \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) ≤ \(\dfrac{-3}{4}\)
c) \(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)
a: ĐKXĐ: x>=3
Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)
=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)
=>\(\dfrac{3}{2}\sqrt{x-3}=3\)
=>\(\sqrt{x-3}=2\)
=>x-3=4
=>x=7(nhận)
b: ĐKXĐ: x>=0
\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)
=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)
=>\(7\sqrt{x}-5< =0\)
=>\(\sqrt{x}< =\dfrac{5}{7}\)
=>0<=x<=25/49
c: ĐKXĐ: x>=5
\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)
=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)
=>\(\dfrac{3}{2}\sqrt{x-5}=3\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
Phương pháp 5. Biến đổi về dạng tổng các bình phương \(A^2+B^2+C^2=0\)
a \(x+y+12=4\sqrt{x}+6\sqrt{y-1}\)
b \(x+y+z+35=2\left(2\sqrt{x+1}+3\sqrt{y+2}+4\sqrt{z+3}\right)\)
c \(9x+17=6\sqrt{8x+1}+4\sqrt{x+3}\)
d \(\sqrt{x}+2\sqrt{x+3}=x+4\)
e\(\sqrt{3-x}+2\sqrt{3x-2}-3=x\)
a.
ĐKXĐ: $x\geq 0; y\geq 1$
PT $\Leftrightarrow (x-4\sqrt{x}+4)+(y-1-6\sqrt{y-1}+9)=0$
$\Leftrightarrow (\sqrt{x}-2)^2+(\sqrt{y-1}-3)^2=0$
Vì $(\sqrt{x}-2)^2; (\sqrt{y-1}-3)^2\geq 0$ với mọi $x\geq 0; y\geq 1$ nên để tổng của chúng bằng $0$ thì:
$\sqrt{x}-2=\sqrt{y-1}-3=0$
$\Leftrightarrow x=4; y=10$
b.
ĐKXĐ: $x\geq -1; y\geq -2; z\geq -3$
PT $\Leftrightarrow x+y+z+35-4\sqrt{x+1}-6\sqrt{y+2}-8\sqrt{z+3}=0$
$\Leftrightarrow [(x+1)-4\sqrt{x+1}+4]+[(y+2)-6\sqrt{y+2}+9]+[(z+3)-8\sqrt{z+3}+16]=0$
$\Leftrightarrow (\sqrt{x+1}-2)^2+(\sqrt{y+2}-3)^2+(\sqrt{z+3}-4)^2=0$
$\Rightarrow \sqrt{x+1}-2=\sqrt{y+2}-3=\sqrt{z+3}-4=0$
$\Rightarrow x=3; y=7; z=13$
c.
ĐKXĐ: $x\geq \frac{-1}{8}$
PT $\Leftrightarrow 9x+17-6\sqrt{8x+1}-4\sqrt{x+3}=0$
$\Leftrightarrow [(8x+1)-6\sqrt{8x+1}+9]+[(x+3)-4\sqrt{x+3}+4]=0$
$\Leftrightarrow (\sqrt{8x+1}-3)^2+(\sqrt{x+3}-2)^2=0$
$\Rightarrow \sqrt{8x+1}-3=\sqrt{x+3}-2=0$
$\Rightarrow x=1$ (thỏa mãn đkxđ)