tìm x biết:
a \(\sqrt{\left(x+1\right)^2}\) = 5
b, 5\(\sqrt{x-9}\) - \(\sqrt{4\left(x-1\right)}\) + \(\sqrt{36\left(x-1\right)}\) -18 = 0
Câu 1: Rút gọn các biểu thức sau:
a. \(\sqrt{36\left(x-5\right)^2}\) với x ≥ 5
b. \(\sqrt{\dfrac{1}{4}\left(1-x\right)^2}\) với x > 1
c. \(\sqrt{x^2\left(2x-4\right)^2}\) với a ≥ 2
d. \(\dfrac{1}{x}\sqrt{x^2\left(1+x\right)^2}\) với x < -1
a) \(\sqrt{36\left(x-5\right)^2}\left(x\ge5\right)=6\left|x-5\right|=6\left(x-5\right)=6x-30\)
b) \(\sqrt{\dfrac{1}{4}\left(1-x\right)^2}\left(x>1\right)=\dfrac{1}{2}\left|1-x\right|=\dfrac{1}{2}\left(x-1\right)=\dfrac{1}{2}x-\dfrac{1}{2}\)
c) \(\sqrt{x^2\left(2x-4\right)^2}\left(x\ge2\right)=\left|x\left(2x-4\right)\right|=x\left(2x-4\right)=2x^2-4x\)
d) \(\dfrac{1}{x}\sqrt{x^2\left(1+x\right)^2}\left(x< -1\right)=\dfrac{1}{x}\left|x\left(1+x\right)\right|=\dfrac{1}{x}x\left(1+x\right)=1+x\)
Bài 3: Tìm x biết:
a) \(\sqrt{3x-2}=4\)
b) \(\sqrt{4x^2+4x+1}-11=5\)
Bài 4: Cho biểu thức
C= \(1\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\) (x > 0, x ≠ 1)
a) Rút gọn C
b) Tìm x để C - 6 < 0
Helpp!!!
Bài 3:
a) \(\sqrt{3x-2}=4\)
⇔\(\sqrt{3x-2}=\sqrt{4^2}\)
⇔\(3x-2=4^2=16\)
\(3x=16+2=18\)
\(x=18:3=6\)
Vậy \(x=6\)
b)\(\sqrt{4x^2+4x+1}-11=5\)
⇔\(\sqrt{\left(2x\right)^2+2\left(2x\right)\cdot1+1^2}-11=5\)
⇔\(\sqrt{\left(2x+1\right)^2}-11=5\)
TH1:
⇔\(\left(2x+1\right)-11=5\)
\(2x+1=5+11=16\)
\(2x=16-1=15\)
\(x=15:2=7,5\)
TH2:
⇔\(\left(2x+1\right)-11=-5\)
\(2x-1=-5+11=6\)
\(2x=6+1=7\)
\(x=7:2=3,5\)
Vậy \(x=\left\{7,5;3,5\right\}\)
(Câu này mình không chắc chắn lắm)
(Học sinh lớp 6 đang làm bài này)
Bài 4:
a: \(C=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+x+\sqrt{x}}{\sqrt{x}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)
b: C-6<0
=>C<6
=>\(2\sqrt{x}< 6\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< x< 9\\x\ne1\end{matrix}\right.\)
Bài 3
a)\(\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\)
Vậy PT có nghiệm x=6
b)\(\sqrt{4x^2+4x+1}-11=5\Leftrightarrow\sqrt{\left(2x+1\right)^2}=16\Leftrightarrow2x+1=16hoặc2x+1=-16\)
+)TH1: \(2x+1=16\Leftrightarrow x=\dfrac{15}{2}\Leftrightarrow x=7,5\)
+)TH2:\(2x+1=-16\Leftrightarrow x=\dfrac{17}{2}\Leftrightarrow x=8,5\)
Bài 4
a)\(C=1\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\Leftrightarrow C=\dfrac{x-1}{\sqrt{x}}\left(\dfrac{x-\sqrt{x}+x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\Leftrightarrow C=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\dfrac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\Leftrightarrow C=\dfrac{2x}{\sqrt{x}}\Leftrightarrow C=2\sqrt{x}\)
\(Vậy\) \(C=2\sqrt{x}\)
* Thực hiện phép tính:
a. \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
b. \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
c. \(\left(\dfrac{6-2\sqrt{2}}{3-\sqrt{2}}\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{2-\sqrt{5}}\)
* Tìm x, biết:
a. \(\sqrt{\left(2x+3\right)^2}=8\)
b. \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)
c. \(\sqrt{9x-9}+1=13\)
bài 1:
a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)
\(=-33\sqrt{2}\)
b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
\(=10-2\sqrt{21}+14\sqrt{21}\)
\(=12\sqrt{21}+10\)
Bài 2:
a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)
\(\Leftrightarrow\left|2x+3\right|=8\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)
b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)
\(\Leftrightarrow4\sqrt{x}=8\)
hay x=4
c: Ta có: \(\sqrt{9x-9}+1=13\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow x-1=16\)
hay x=17
Ai dậy r giúp vs :33 1 câu cx đc nhé :v toàn giải pt hết nhé
1) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}.\)
2) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+18-\left(x+26\right)\sqrt{x-1}\)
3) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
4) \(\left(x+17\right)\sqrt{4-x}+\left(20-x\right)\sqrt{x+1}-9\sqrt{4-x}.\sqrt{x+1}=36\)
Câu 1 là \(\left(8x-4\right)\sqrt{x}-1\) hay là \(\left(8x-4\right)\sqrt{x-1}\)?
Câu 1:ĐK \(x\ge\frac{1}{2}\)
\(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
<=> \(\left(4x^2-3x-1\right)+4\left(2x-1\right)\sqrt{x}-2\sqrt{\left(2x-1\right)\left(x+3\right)}\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}\left(2\sqrt{x\left(2x-1\right)}-\sqrt{x+3}\right)=0\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{8x^2-4x-x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=>\(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{\left(x-1\right)\left(8x+3\right)}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=> \(\left(x-1\right)\left(4x+1+2\sqrt{2x-1}.\frac{8x+3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}\right)=0\)
Với \(x\ge\frac{1}{2}\)thì \(4x+1+2\sqrt{2x-1}.\frac{8x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}>0\)
=> \(x=1\)(TM ĐKXĐ)
Vậy x=1
câu 2 ĐK \(x\ge1\)
\(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+18-\left(x+26\right)\sqrt{x-1}=0\)
<=> \(\left(5x+8\right)\left(\sqrt{2x-1}-1\right)+7x\left(\sqrt{x+3}-2\right)+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)
<=>\(\left(5x+8\right).\frac{2x-2}{\sqrt{2x-1}+1}+7x.\frac{x+3-4}{\sqrt{x+3}+2}+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)
<=> \(\sqrt{x-1}\left(\frac{2\left(5x+8\right)\sqrt{x-1}}{\sqrt{2x-1}+1}+\frac{7x\sqrt{x-1}}{\sqrt{x+3}+2}+\left(x+26\right)+10\sqrt{x-1}\right)=0\)
Với \(x\ge1\)thì cái trong ngoặc >0
=> \(x=1\)
Vậy x=1
\(\left(1\right)\sqrt{x^2-9}-2\sqrt{x-3}=0\)
\(\left(2\right)\sqrt{4x+1}-\sqrt{3x-4}=1\)
\(\left(3\right)\sqrt{x^2-10x+25}=5-x\)
\(\left(4\right)\sqrt{x^2-8x+16}=x+2\)
1:
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-2\right)=0\)
=>x-3=0 hoặc \(\sqrt{x+3}=2\)
=>x=3 hoặc x+3=4
=>x=1(loại) hoặc x=3(nhận)
2:
\(\Leftrightarrow\left(\sqrt{4x+1}-\sqrt{3x-4}\right)^2=1\)
=>\(4x-1+3x-4-2\sqrt{\left(4x+1\right)\left(3x-4\right)}=1\)
=>\(\sqrt{4\left(4x+1\right)\left(3x-4\right)}=7x-6\)
=>4(12x^2-16x+3x-4)=(7x-6)^2
=>49x^2-84x+36=48x^2-52x-16
=>-84x+36=-52x-16
=>-32x=-52
=>x=13/8
3: =>\(\sqrt{\left(x-5\right)^2}=5-x\)
=>|x-5|=5-x
=>x-5<=0
=>x<=5
4: \(\Leftrightarrow\left|x-4\right|=x+2\)
=>\(\left\{{}\begin{matrix}x>=-2\\\left(x-4\right)^2=\left(x+2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\x^2-8x+16=x^2+4x+4\end{matrix}\right.\)
=>x>=-2 và -8x+16=4x+4
=>x=1
Tìm x, biết:
a) \(\dfrac{-3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)
b) \(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)
a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)
=>\(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}x+\dfrac{1}{2}\)
=>\(-\dfrac{3}{2}x-\dfrac{1}{2}x=\dfrac{1}{2}-\dfrac{1}{4}\)
=>\(-2x=\dfrac{1}{4}\)
=>\(2x=-\dfrac{1}{4}\)
=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)
b: ĐKXĐ: x>=0
\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)
=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
\(\sqrt{\left(2x+3\right)^2}=5\)
\(\sqrt{9.\left(x-2\right)^2}=18\)
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
\(\sqrt{4.\left(x-3\right)^2}=8\)
\(\sqrt{4x^2+12x+9}=5\)
\(\sqrt{5x-6}-3=0\)
a: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{9\left(x-2\right)^2}=18\)
=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)
=>\(3\cdot\left|x-2\right|=18\)
=>\(\left|x-2\right|=6\)
=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
=>\(4\sqrt{x-2}=40\)
=>\(\sqrt{x-2}=10\)
=>x-2=100
=>x=102(nhận)
d: ĐKXĐ: \(x\in R\)
\(\sqrt{4\left(x-3\right)^2}=8\)
=>\(\sqrt{\left(2x-6\right)^2}=8\)
=>|2x-6|=8
=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
e: ĐKXĐ: \(x\in R\)
\(\sqrt{4x^2+12x+9}=5\)
=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)
=>\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
f: ĐKXĐ:x>=6/5
\(\sqrt{5x-6}-3=0\)
=>\(\sqrt{5x-6}=3\)
=>\(5x-6=3^2=9\)
=>5x=6+9=15
=>x=15/5=3(nhận)
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
Tìm các giá trị của biến để :
\(9.x^2-36=0\)
\(\left(x-1\right).\left(x+1\right).\left(x^2+\dfrac{1}{2}\right)=0\)
\(\left|x+4\right|+5=0\)
\(\sqrt{2.x}-3-1=0\)
Mn làm cho mik dò điii
a/ \(\Leftrightarrow9x^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-6\end{matrix}\right.\)
\(\Leftrightarrow x=\pm2\)
b/ \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\) (do \(x^2+\dfrac{1}{2}>0\))
\(\Leftrightarrow x=\pm1\)
c/ Có \(\left|x+4\right|\ge0\forall x\)
=> \(\left|x+4\right|+5\ge5>0\forall x\)
\(\Rightarrow\left|x+4\right|+5=0\left(vô-lí\right)\)
\(\Rightarrow x\in\varnothing\)
d/ \(\sqrt{2x}-3-1=0\)
\(\Leftrightarrow\sqrt{2x}=4\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)