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a: \(\Leftrightarrow x+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
hay \(x\in\left\{-2;-4;-1;-5;0;-6;1;-7;3;-9;9;-15\right\}\)
\(a)\)\(\sqrt{9\left(x-1\right)^2}-12=0\)
\(\Leftrightarrow\)\(\sqrt{\left[3\left(x-1\right)\right]^2}=12\)
\(\Leftrightarrow\)\(\left|3\left(x-1\right)\right|=12\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3\left(x-1\right)=12\\3\left(x-1\right)=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
Vậy \(x=-3\) hoặc \(x=5\)
\(b)\) ĐKXĐ : \(x\le\frac{3}{2}\)
\(\sqrt{2-3x}=10\)
\(\Leftrightarrow\)\(2-3x=100\)
\(\Leftrightarrow\)\(3x=-98\)
\(\Leftrightarrow\)\(x=\frac{-98}{3}\) ( không thỏa mãn )
Vậy không có x thỏa mãn đề bài
\(c)\) ĐKXĐ : \(x>2\) hoặc \(x\le-2\)
\(\sqrt{x^2-4}=\sqrt{2-x}\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2\right)\left(x+2\right)}-\sqrt{2-x}=0\)
\(\Leftrightarrow\)\(\sqrt{2-x}.\sqrt{-x-2}-\sqrt{2-x}=0\)
\(\Leftrightarrow\)\(\sqrt{2-x}\left(\sqrt{-x-2}-1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}-\sqrt{2-x}=0\\\sqrt{-x-2}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=-3\left(nhan\right)\end{cases}}}\)
Vậy \(x=-3\)
Chúc bạn học tốt ~
PS : mới lớp 8 :v làm sai thì thông cảm
c) \(\sqrt{\left(x-2\right)^2}=10\)
\(x-2=10\)
\(x=12\)
d) \(\sqrt{9x^2-6x+1}=15\)
\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)
\(\sqrt{\left(3x-1\right)^2}=15\)
\(3x-1=15\)
\(3x=16\)
\(x=\dfrac{16}{3}\)
a) \(đk:x\ge0\)
\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)
\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)
b) \(đk:x\ge-2\)
\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)
\(\Leftrightarrow13\sqrt{x+2}=26\)
\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)
c) \(pt\Leftrightarrow\left|x-2\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)
\(\Leftrightarrow\left|3x-1\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)
e) \(đk:x\ge\dfrac{8}{3}\)
\(pt\Leftrightarrow3x+4=9x^2-48x+64\)
\(\Leftrightarrow9x^2-51x+60=0\)
\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
Câu 2:
a: Ta có: \(P=3x-\sqrt{x^2-10x+25}\)
\(=3x-\left|x-5\right|\)
\(=\left[{}\begin{matrix}3x-x+5=2x+5\left(x\ge5\right)\\3x+x-5=4x-5\left(x< 5\right)\end{matrix}\right.\)
b: Vì x=2<5 nên \(P=4\cdot2-5=8-5=3\)
a: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(x+3\right)^2}=12\)
=>\(\left|x+3\right|=12\)
=>\(\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)
b: ĐKXĐ: x>=1
\(\sqrt{25x-25}-\sqrt{9x-9}=10\)
=>\(5\sqrt{x-1}-3\sqrt{x-1}=10\)
=>\(2\sqrt{x-1}=10\)
=>x-1=25
=>x=26(nhận)
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
a) \(\sqrt{x-3}>2\left(đk:x\ge3\right)\)
\(\Leftrightarrow x-3>4\Leftrightarrow x>7\)
b) \(\sqrt{36x^2-12x+1}=5\)
\(\Leftrightarrow\sqrt{\left(6x-1\right)^2}=5\)
\(\Leftrightarrow\left|6x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-\dfrac{2}{3}\end{matrix}\right.\)
c) \(\sqrt{9\left(5x^2-2x+16\right)}=3x+12\left(đk:x\ge-4\right)\)
\(\Leftrightarrow9\left(5x^2-2x+16\right)=9x^2+72x+144\)
\(\Leftrightarrow36x^2-90x=0\)
\(\Leftrightarrow18x\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
[ ( 3x - 9 ) : 10 ] + 2 = 12
( 3x - 9 ) : 10 = 10
3x - 9 = 100
3x = 109
x = 109/3
Hỏi chơi thôi phải không
đáp án chính xác là 109/3