Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)√x2−9 - 3√x−3 =0
<=> (√x-3)(√x+3)-3√x-3=0
<=> (√x-3)(√x+3-3)=0
<=> (√x-3)√x=0
<=> √x-3=0
<=>x=9
b)√4x2−12x+9=x - 3
<=> √(2x -3)2 =x-3
<=> 2x-3=x-3
<=>2x-x=-3+3
<=>x=0
c)√x2+6x+9=3x-1
<=> √(x+3)2 =3x-1
<=> x+3=3x-1
<=> -2x=-4
<=> x=2
Nhớ cho mình 1 tim nha bạn
Sau em nên gõ các kí hiệu toán học ở phần Σ để mọi người dễ dàng đọc hơn nhé.
a) \(\sqrt{x^2-9}-\sqrt{4x-12}=0\) ĐK: \(x\ge3\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-2\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy x = 3
b) \(\sqrt{1-x}+\sqrt{x}=1\) ĐK: \(0\le x\le1\)
\(\Leftrightarrow1-x+x+2\sqrt{x\left(1-x\right)}=1\)
\(\Leftrightarrow x\left(1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) (Nhận)
c) \(\sqrt{x+3}+\sqrt{x+8}=5\) ĐK: \(x\ge-3\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+3}\ge0\\b=\sqrt{x+8}\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=5\\b^2-a^2=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\b-a=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)
\(\Leftrightarrow x=1\) (Nhận)
d) \(\sqrt{16-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}\) ĐK: \(-\dfrac{1}{2}\le x\le0\)
\(\Leftrightarrow4\sqrt{1-2x}-2\sqrt{3x}=\sqrt{3x}+3\sqrt{1-2x}\)
\(\Leftrightarrow\sqrt{1-2x}=3\sqrt{3x}\)
\(\Leftrightarrow1-2x=27x\)
\(\Leftrightarrow x=\dfrac{1}{29}\) (Nhận)
1: Ta có: \(\sqrt{4x^2-12x+9}=3-2x\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(3-2x\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(3-2x\right)^2=0\)
\(\Leftrightarrow\left[\left(2x-3\right)-\left(3-2x\right)\right]\left[\left(2x-3\right)+\left(3-2x\right)\right]=0\)
\(\Leftrightarrow\left(2x-3-3+2x\right)\left(2x-3+3-2x\right)=0\)
\(\Leftrightarrow\left(4x-6\right)\cdot0=0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
2) Ta có: \(\sqrt{x^2-2\cdot\sqrt{2}\cdot x+2}=\sqrt{9-4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{8-2\cdot2\sqrt{2}\cdot1+1}-\sqrt{1+2\cdot1\cdot\sqrt{2}+2}\)
\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\left|\sqrt{8}-1\right|-\left|1+\sqrt{2}\right|\)
\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{8}-1-1-\sqrt{2}\)
\(\Leftrightarrow\left|x-\sqrt{2}\right|=\sqrt{2}-2\)(*)
Trường hợp 1: \(x\ge\sqrt{2}\)
(*)\(\Leftrightarrow x-\sqrt{2}=\sqrt{2}-2\)
\(\Leftrightarrow x-\sqrt{2}-\sqrt{2}+2=0\)
\(\Leftrightarrow x-2\sqrt{2}+2=0\)
\(\Leftrightarrow x=2\sqrt{2}-2\)(loại)
Trường hợp 2: \(x< \sqrt{2}\)
(*)\(\Leftrightarrow\sqrt{2}-x=\sqrt{2}-2\)
\(\Leftrightarrow\sqrt{2}-x-\sqrt{2}+2=0\)
\(\Leftrightarrow2-x=0\)
hay x=2(loại)
Vậy: S=∅
\(1.\sqrt{16-8x+x^2}=4-x\)
\(\sqrt{\left(4-x\right)^2}=4-x\)
\(4-x-4+x=0\)
= 0 phương trình vô nghiệm.
\(2.\sqrt{4x^2-12x+9}=2x-3\)
\(\)\(\sqrt{\left(2x-3\right)^2}=2x-3\)
\(2x-3-2x+3=0\)
= 0 phương trình vô nghiệm.
a: Ta có: \(\sqrt{16-8x+x^2}=4-x\)
\(\Leftrightarrow\left|4-x\right|=4-x\)
hay \(x\le4\)
b: Ta có: \(\sqrt{4x^2-12x+9}=2x-3\)
\(\Leftrightarrow\left|2x-3\right|=2x-3\)
hay \(x\ge\dfrac{3}{2}\)
\(1)\) ĐKXĐ : \(x\ge3\)
\(\sqrt{x^2-4x+3}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x^2-4x+4\right)-1}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2\right)^2-1}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2-1\right)\left(x-2+1\right)}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-3\right)\left(x-1\right)}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{x-1}\left(\sqrt{x-3}+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x-3}+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x\in\left\{\varnothing\right\}\end{cases}}}\)
Vậy \(x=1\)
\(2)\)\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)
\(\Leftrightarrow\)\(\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-3\right)^2}=10\)
\(\Leftrightarrow\)\(\left|x-1\right|-\left|x-3\right|=10\)
+) Với \(\hept{\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\ge3\end{cases}\Leftrightarrow}x\ge3}\) ta có :
\(x-1-x+3=10\)
\(\Leftrightarrow\)\(0=8\) ( loại )
+) Với \(\hept{\begin{cases}x-1< 0\\x-3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 1\\x< 3\end{cases}\Leftrightarrow}x< 1}\) ta có :
\(1-x+x-3=10\)
\(\Leftrightarrow\)\(0=12\) ( loại )
Vậy không có x thỏa mãn đề bài
Chúc bạn học tốt ~
PS : mới lp 8 sai đừng chửi nhé :v
a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
=>x-2=16
hay x=18
b: \(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)
c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
\(\Leftrightarrow4\sqrt{x-2}=40\)
=>x-2=100
hay x=102
d: =>5x-6=9
hay x=3
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
6) \(\sqrt{x^2-4x+1}=x\left(x\ge0\right)\)
\(\Leftrightarrow x^2-4x+1=x^2\)
\(\Leftrightarrow x^2-x^2=4x-1\)
\(\Leftrightarrow4x=1\)
\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
8) \(\sqrt{x^2-x-6}=\sqrt{x-3}\left(x\ge3\right)\)
\(\Leftrightarrow x^2-x-6=x-3\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
9) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\left(x\ge1\right)\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=1+1\)
\(\Leftrightarrow x=2\left(tm\right)\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
a) \(\sqrt{1-x}=\sqrt[3]{8}\) ( ĐK: \(x\le1\) )
\(\Leftrightarrow\sqrt{1-x}=2\)
\(\Leftrightarrow1-x=4\)
\(\Leftrightarrow x=-3\) ( Thỏa mãn )
b) \(\sqrt{4x^2-12x+9}=x+1\) ( ĐK : \(x\ge-1\) )
\(\Leftrightarrow\sqrt{\left(2x\right)^2-2.2x.3+3^2}=x+1\)
\(\Leftrightarrow\sqrt{\left(2x-3\right)^2}=x+1\)
\(\Leftrightarrow\left|2x-3\right|=x+1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=x+1\\3-2x=x+1\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{2}{3}\end{cases}}\) ( Thỏa mãn )
c) \(x+\sqrt{x}-2=0\) ( ĐK : \(x\ge0\) )
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\sqrt{x}-1=0\)
\(\Leftrightarrow x=1\) ( Thỏa mãn )
+) ĐKXĐ : \(x\le1\)
\(\sqrt{1-x}=\sqrt[3]{8}\)
\(\Leftrightarrow\sqrt{1-x}=2\)
\(\Leftrightarrow1-x=4\)
\(\Leftrightarrow x=-3\left(TM\right)\)
+) \(\sqrt{4x^2-12x+9}=x+1\)
\(\Leftrightarrow\sqrt{\left(2x-3\right)^2}=x+1\)
\(\Leftrightarrow\left|2x-3\right|=x+1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=x+1\left(x\ge\frac{3}{2}\right)\\2x-3=-x-1\left(x< \frac{3}{2}\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-x=3+1\\2x+x=3-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\3x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{2}{3}\end{cases}\left(TM\right)}}\)
+) ĐKXĐ : \(x\ge0\)
\(x+\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=2\)
+) \(\hept{\begin{cases}\sqrt{x}=1\\\sqrt{x}+1=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=1\end{cases}\Leftrightarrow}x=1\left(TM\right)}\)
+) \(\hept{\begin{cases}\sqrt{x}=2\\\sqrt{x}+1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\sqrt{2}\\x=0\end{cases}}}\left(TM\right)\)