Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1 )
a)\(3\sqrt{\frac{1}{3}}-\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{2}\)
b)\(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}=\left(\sqrt{3}+1\right)-\left|1-\sqrt{3}\right|=\left(\sqrt{3}+1\right)-\sqrt{3}+1=2\)
Bài 2)
a)\(\sqrt{36x^2-12x+1}=5\)
\(\Leftrightarrow36x^2-12x+1=25\)
\(\Leftrightarrow36x^2-12x+1=25\)
\(\Leftrightarrow\left(6x\right)^2-2.6x+1=25\)
\(\Leftrightarrow\left(6x-1\right)^2=25\)
\(\Rightarrow6x-1=5\)
\(\Leftrightarrow6x=6\)
\(\Rightarrow x=1\)
b)\(\sqrt{x-5}-2\sqrt{4x-20}-\frac{1}{3}\sqrt{9x-45}=12\)
\(\Leftrightarrow\sqrt{x-5}-2\sqrt{4.\left(x-5\right)}-\frac{1}{3}\sqrt{9.\left(x-5\right)}=12\)
\(\Leftrightarrow\sqrt{x-5}-4\sqrt{\left(x-5\right)}-\sqrt{\left(x-5\right)}=12\)
\(\Leftrightarrow-4\sqrt{\left(x-5\right)}=12\)
\(\Rightarrow\)ko tồn tại giá trị nào của x trong biểu thức này
P/s tham khảo nha
1a) \(3\sqrt{\frac{1}{3}}-\frac{1}{\sqrt{3}+\sqrt{2}}\)
=\(3\sqrt{\frac{3}{3^2}}-\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
=\(3\frac{\sqrt{3}}{\sqrt{3^2}}-\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}\)
=\(3\frac{\sqrt{3}}{3}-\frac{\sqrt{3}-\sqrt{2}}{3-2}\)
=\(\sqrt{3}-\left(\sqrt{3}-\sqrt{2}\right)\)
=\(\sqrt{3}-\sqrt{3}+\sqrt{2}\)=\(\sqrt{2}\)
b)\(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
=\(|\sqrt{3}+1|-|1-\sqrt{3}|\)
=\(\sqrt{3}+1-\left(-\left(1-\sqrt{3}\right)\right)\)
=\(\sqrt{3}+1+1-\sqrt{3}\)
=\(1+1\)=\(2\)
2) a) \(\sqrt{36x^2-12x+1}=5\)
<=>\(\sqrt{\left(6x\right)^2-2.6x.1+1^2}=5\)
<=>\(\sqrt{\left(6x-1\right)^2}=5\)
<=>\(|6x-1|=5\)
Nếu \(6x-1>=0\)=> \(6x>=1\)=>\(x>=\frac{1}{6}\)
Nên \(|6x-1|=6x-1\)
Ta có \(|6x-1|=5\)
<=> \(6x-1=5\)
<=> \(6x=6\)
<=> \(x=1\)(thỏa)
Nếu \(6x-1< 0\)=> \(6x< 1\)=>\(x< \frac{1}{6}\)
Nên \(|6x-1|=-\left(6x-1\right)=1-6x\)
Ta có \(|6x-1|=5\)
<=> \(1-6x=5\)
<=> \(-6x=4\)
<=> \(x=\frac{4}{-6}=\frac{-2}{3}\)(thỏa)
Vậy \(x=1\)và \(x=\frac{-2}{3}\)
b) \(\sqrt{x-5}-2\sqrt{4x-20}-\frac{1}{3}\sqrt{9x-45}=12\)
<=>\(\sqrt{x-5}-2\sqrt{4\left(x-5\right)}-\frac{1}{3}\sqrt{9\left(x-5\right)}=12\)
<=>\(\sqrt{x-5}-2.2\sqrt{x-5}-\frac{1}{3}.3\sqrt{x-5}=12\)
<=>\(\sqrt{x-5}-4\sqrt{x-5}-\sqrt{x-5}=12\)
<=>\(-4\sqrt{x-5}=12\)
<=> \(\sqrt{x-5}=-3\)
<=> \(\left(\sqrt{x-5}\right)^2=\left(-3\right)^2\)
<=>\(x-5=9\)
<=>\(x=14\)
Vậy x=14
Kết bạn với mình nhá
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
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
a) ĐK: \(x\ge0\)
PT \(\Leftrightarrow\sqrt{4x}\left(\dfrac{3}{4}-1-\dfrac{1}{4}\right)+5=0\)
\(\Leftrightarrow2\sqrt{x}.\left(-\dfrac{1}{2}\right)+5=0\)
\(\Leftrightarrow x=25\) (thỏa)
Vậy \(x=25\)
b) Đk: \(x\le3\)
PT \(\Leftrightarrow\sqrt{3-x}-\sqrt{9\left(3-x\right)}+\dfrac{5}{4}\sqrt{16\left(3-x\right)}=6\)
\(\Leftrightarrow\sqrt{3-x}\left(1-\sqrt{9}+\dfrac{5}{4}.\sqrt{16}\right)=6\)
\(\Leftrightarrow\sqrt{3-x}=2\Leftrightarrow x=-1\) (thỏa)
Vậy \(x=-1\)
2:
a:
Sửa đề: \(P=\left(\dfrac{2}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{2}{\sqrt{1-a^2}}+1\right)\)
\(P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{1-a^2}}{\sqrt{1-a^2}}\)
\(=\dfrac{2+\sqrt{1-a^2}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{1-a^2}}{2+\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}\)
\(=\sqrt{1-a}\)
b: Khi a=24/49 thì \(P=\sqrt{1-\dfrac{24}{49}}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\)
c: P=2
=>1-a=4
=>a=-3
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)
ĐKXĐ: \(x> \frac{-5}{7}\)
Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)
\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)
Vậy......
b) ĐKXĐ: \(x\geq 5\)
\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)
(hoàn toàn thỏa mãn)
Vậy..........
c) ĐK: \(x\in \mathbb{R}\)
Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)
\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
Khi đó:
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)
\(\Leftrightarrow 7-a^2+6a=0\)
\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)
\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)
\(\Rightarrow 6x^2-12x+7=a^2=49\)
\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)
\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)
(đều thỏa mãn)
Vậy..........
1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)
2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)
3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)
4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)
a) Ta có: \(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=2\sqrt{5}-6\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=-4\sqrt{5}+15\sqrt{2}\)
b) Ta có: \(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)
\(=4\left(\sqrt{3}-1\right)+2\cdot2\sqrt{3}+\dfrac{4}{\sqrt{2}}\)
\(=4\sqrt{3}-4+4\sqrt{3}+2\sqrt{2}\)
\(=8\sqrt{3}+2\sqrt{2}-4\)
c) Ta có: \(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3
=6
d) Ta có: \(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)
\(=2-\sqrt{3}+2+\sqrt{3}\)
=4
1)
a) \(3\sqrt{\dfrac{1}{3}}-\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
\(=\sqrt{3}-\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=\sqrt{2}\)
b) \(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{3}+1\right|-\left|1-\sqrt{3}\right|\)
\(=\sqrt{3}+1-\sqrt{3}+1\)
\(=2\)