Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A) \(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\)
\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\dfrac{\sqrt{x-1}}{3}-\sqrt{x-1}=6\)
\(\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\sqrt{x-1}-\sqrt{x-1}=6\)
\(\Leftrightarrow\dfrac{3}{2}\sqrt{x-1}=6\)
\(\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\)
\(\Leftrightarrow x=17\)
Vậy, x=17
A: \(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)
=>5/2*căn x-1-căn x-1=6
=>3/2*căn x-1=6
=>căn x-1=4
=>x-1=16
=>x=17
B:
a: ĐKXĐ: x>=0; x<>1
b: Sửa đề: \(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\)
=căn x-1+x-căn x+1
=x
\(\sqrt{25x-25}-\dfrac{15}{2}\cdot\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\) (1)
\(\Leftrightarrow\sqrt{25\left(x-1\right)}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{25}\sqrt{x-1}-\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}\)
\(\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}\)
\(\Leftrightarrow\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}\)
\(\Leftrightarrow5\sqrt{x-1}=12+2\sqrt{x-1}\)
\(\Leftrightarrow5\sqrt{x-1}-2\sqrt{x-1}=12\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
\(\Leftrightarrow x=16+1\)
\(\Leftrightarrow x=17\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{17\right\}\)
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
1) \(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}+1-\sqrt{3}+1=2\)
2) \(\dfrac{3}{5}\sqrt{25x-50}-\sqrt{x-2}=6\left(đk:x\ge2\right)\)
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{x-2}=6\)
\(\Leftrightarrow2\sqrt{x-2}=6\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\)
\(a.\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)( x lớn hơn hoặc =1)
\(\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}\)+2=0
\(\sqrt{x-1}\left(1+\sqrt{4}-\sqrt{25}\right)=-2\)
\(\sqrt{x-1}\left(1+2-5\right)=-2\)
\(\sqrt{x-1}.\left(-2\right)=-2\)
\(\sqrt{x-1}=-2.2\)
\(\sqrt{x-1}-4\)(ko thỏa mãn)
b)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9\left(x-1\right)}+24\dfrac{\sqrt{x-1}}{8}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.3\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\left(\dfrac{1}{2}-\dfrac{9}{2}+3\right)\sqrt{x-1}=-17\)
\(7\sqrt{x-1}=-17\)
\(\sqrt{x-1}=-\dfrac{17}{7}\)(ko thỏa mãn căn bậc 2 ko có số âm)
a: Ta có: \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow x-1=1\)
hay x=2
\(d,ĐK:x\ge0\\ PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+\dfrac{3}{2}\cdot2\sqrt{x-1}-\dfrac{2}{5}\cdot5\sqrt{x-1}=4\\ \Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\\ \Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\\ f,ĐK:x\ge5\\ PT\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\\ \Leftrightarrow2\sqrt{x-5}=6\Leftrightarrow\sqrt{x-5}=3\\ \Leftrightarrow x-5=9\Leftrightarrow x=14\left(tm\right)\)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
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) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
\(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\left(x\ge1\right)\)
\(< =>5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)
\(< =>30\sqrt{x-1}-15\sqrt{x-1}=36+6\sqrt{x-1}\)
\(< =>9\sqrt{x-1}=36\\ < =>\sqrt{x-1}=4\\ < =>x-1=16\\ < =>x=17\left(tm\right)\)
\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{1}{3}\sqrt{x-1}-\sqrt{x-1}=6\)
=>\(1.5\cdot\sqrt{x-1}=6\)
=>\(\sqrt{x-1}=4\)
=>x-1=16
=>x=17