Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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: \(\Leftrightarrow\sqrt{6}\left(x+1\right)=5\sqrt{6}\)
=>x+1=5
=>x=4
b: =>x^2/10=1,1
=>x^2=11
=>x=căn 11 hoặc x=-căn 11
c: =>(4x+3)/(x+1)=9 và (4x+3)/(x+1)>=0
=>4x+3=9x+9
=>-5x=6
=>x=-6/5
d: =>(2x-3)/(x-1)=4 và x-1>0 và 2x-3>=0
=>2x-3=4x-4 và x>=3/2
=->-2x=-1 và x>=3/2
=>x=1/2 và x>=3/2
=>Ko có x thỏa mãn
e: Đặt căn x=a(a>=0)
PT sẽ là a^2-a-5=0
=>\(\left[{}\begin{matrix}a=\dfrac{1+\sqrt{21}}{2}\left(nhận\right)\\a=\dfrac{1-\sqrt{21}}{2}\left(loại\right)\end{matrix}\right.\)
=>x=(1+căn 21)^2/4=(11+căn 21)/2
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 $\Leftrightarrow x+y+z+35-4\sqrt{x+1}-6\sqrt{y+2}-8\sqrt{z+3}=0$
$\Leftrightarrow [(x+1)-4\sqrt{x+1}+4]+[(y+2)-6\sqrt{y+2}+9]+[(z+3)-8\sqrt{z+3}+16]=0$
$\Leftrightarrow (\sqrt{x+1}-2)^2+(\sqrt{y+2}-3)^2+(\sqrt{z+3}-4)^2=0$
$\Rightarrow \sqrt{x+1}-2=\sqrt{y+2}-3=\sqrt{z+3}-4=0$
$\Rightarrow x=3; y=7; z=13$
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: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
Câu a:
ĐKXĐ:...........
\(\sqrt{x^2-x+9}=2x+1\)
\(\Rightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-x+9=(2x+1)^2=4x^2+4x+1\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+5x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x(x-1)+8(x-1)=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (x-1)(3x+8)=0\end{matrix}\right.\Rightarrow x=1\)
Vậy.....
Câu b:
ĐKXĐ:.........
Ta có: \(\sqrt{5x+7}-\sqrt{x+3}=\sqrt{3x+1}\)
\(\Rightarrow (\sqrt{5x+7}-\sqrt{x+3})^2=3x+1\)
\(\Leftrightarrow 5x+7+x+3-2\sqrt{(5x+7)(x+3)}=3x+1\)
\(\Leftrightarrow 3(x+3)=2\sqrt{(5x+7)(x+3)}\)
\(\Leftrightarrow \sqrt{x+3}(3\sqrt{x+3}-2\sqrt{5x+7})=0\)
Vì \(x\geq -\frac{7}{5}\Rightarrow \sqrt{x+3}>0\). Do đó:
\(3\sqrt{x+3}-2\sqrt{5x+7}=0\)
\(\Rightarrow 9(x+3)=4(5x+7)\)
\(\Rightarrow 11x=-1\Rightarrow x=\frac{-1}{11}\) (thỏa mãn)
Vậy..........
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
a) ĐKXĐ: \(x\geq -3\)
Ta có: \(\sqrt{x+3}=1+\sqrt{2}\)
\(\Rightarrow x+3=(1+\sqrt{2})^2\)
\(\Leftrightarrow x+3=1+2+2\sqrt{2}=3+2\sqrt{2}\)
\(\Leftrightarrow x=2\sqrt{2}\) (thỏa mãn)
Vậy \(x=2\sqrt{2}\)
b) ĐK: \(x\geq 0\)
Có: \(\sqrt{10+\sqrt{5x}}=\sqrt{6}+2\)
\(\Rightarrow 10+\sqrt{5x}=(\sqrt{6}+2)^2=6+4+4\sqrt{6}\)
\(\Leftrightarrow \sqrt{5x}=4\sqrt{6}=\sqrt{96}\)
\(\Leftrightarrow x=\frac{96}{5}\) (thỏa mãn)
Vậy.....
c) ĐK: \(x\geq 4\)
Ta có: \(\sqrt{x^2-16}-\sqrt{x-4}=0\)
\(\Leftrightarrow \sqrt{(x-4)(x+4)}-\sqrt{x-4}=0\)
\(\Leftrightarrow \sqrt{x-4}(\sqrt{x+4}-1)=0\)
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-4}=0\\ \sqrt{x+4}=1\end{matrix}\right. \Leftrightarrow \left[\begin{matrix} x=4\\ x=-3\end{matrix}\right.\) (loại $x=-3$ vì $x\geq 4$)
Vậy \(x=4\)
d) ĐK: \(x\ge 0\)
Ta có: \(x-6\sqrt{x}+5=0\)
\(\Leftrightarrow (x-\sqrt{x})-5(\sqrt{x}-1)=0\)
\(\Leftrightarrow \sqrt{x}(\sqrt{x}-1)-5(\sqrt{x}-1)=0\)
\(\Leftrightarrow (\sqrt{x}-5)(\sqrt{x}-1)=0\)
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x}-5=0\\ \sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=25\\ x=1\end{matrix}\right.\) (đều t/m)
e) ĐK: \(x\geq 3\)
\(\sqrt{x-3}\geq 7\)
\(\Leftrightarrow x-3\geq 49\)
\(\Leftrightarrow x\geq 52\). Kết hợp với ĐK suy ra \(x\geq 52\)
f) ĐK: \(x\geq -1\)
Ta có: \(\sqrt{x+1}\leq 3\)
\(\Leftrightarrow x+1\leq 9\)
\(\Leftrightarrow x\leq 8\)
Kết hợp với ĐK suy ra \(-1\leq x\leq 8\)