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a: \(\left(2x-3\right)^2=\left|3-2x\right|\)
=>\(\left\{{}\begin{matrix}\left|2x-3\right|>=0\\\left(2x-3\right)^2=\left(2x-3\right)\end{matrix}\right.\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)=0\)
=>\(\left(2x-3\right)\left(2x-3-1\right)=0\)
=>\(\left(2x-3\right)\left(2x-4\right)=0\)
=>\(\left[{}\begin{matrix}2x-3=0\\2x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
b: \(\left(x-1\right)^2+\left(2x-1\right)^2=0\)
=>\(x^2-2x+1+4x^2-4x+1=0\)
=>\(5x^2-6x+2=0\)
\(\Delta=\left(-6\right)^2-4\cdot5\cdot2=36-20\cdot2=-4< 0\)
=>Phương trình vô nghiệm
c: ĐKXĐ: x>=0
\(x-2\sqrt{x}=0\)
=>\(\sqrt{x}\cdot\sqrt{x}-2\cdot\sqrt{x}=0\)
=>\(\sqrt{x}\left(\sqrt{x}-2\right)=0\)
=>\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
d: \(\left(x-1\right)^2+\dfrac{1}{7}=0\)
mà \(\left(x-1\right)^2+\dfrac{1}{7}>=\dfrac{1}{7}>0\forall x\)
nên \(x\in\varnothing\)
Rút gọn: −3x3+2x2+x
Hệ số cao nhất : -3
Hệ số tự do là: 0
Không có đáp án đúng
\(P\left(x\right)=-3x^3+2x^2+x\)
=>Không có câu nào đúng
Lời giải:
a.
$|2x-5|=12-3x$
Nếu $x\geq \frac{5}{2}$ thì $2x-5=12-3x$
$\Leftrightarrow x=3,4$ (thỏa mãn)
Nếu $x< \frac{5}{2}$ thì: $5-2x=12-3x$
$\Leftrightarrow x=7$ (loại)
Vậy......
b.
$4x=|x+1|+|x+2|+|x+3|\geq 0$
$\Rightarrow x\geq 0$
Do đó: $|x+1|+|x+2|+|x+3|=(x+1)+(x+2)+(x+3)=3x+6$
Vậy: $3x+6=4x$
$\Leftrightarrow x=6$ (thỏa mãn)
c.
$|x^2+|x+2||=x^2+3$
$\Leftrightarrow x^2+|x+2|=x^2+3$
$\Leftrightarrow |x+2|=3$
$\Leftrightarrow x+2=3$ hoặc $x+2=-3$
$\Leftrightarrow x=1$ hoặc $x=-5$
d.
$|x^2-3|=6$
$\Leftrightarrow x^2-3=6$ hoặc $x^2-3=-6$
$\Leftrightarrow x^2=9$ (chọn) hoặc $x^2=-3< 0$ (loại)
$\Leftrightarrow x=\pm 3$
a, \(\dfrac{x}{2}+\dfrac{3x}{5}=-\dfrac{3}{2}\Rightarrow5x+6x=-15\Leftrightarrow x=-\dfrac{15}{11}\)
b, TH1 : \(\dfrac{2}{3}x-\dfrac{4}{7}=0\Leftrightarrow x=\dfrac{6}{7}\);TH2 : \(\dfrac{1}{2}-\dfrac{3}{7x}=0\Rightarrow7x-6=0\Leftrightarrow x=\dfrac{6}{7}\)
c, TH1 : \(\dfrac{4}{5}-2x=0\Leftrightarrow x=\dfrac{4}{5}:2=\dfrac{2}{5}\)
TH2 : \(\dfrac{1}{3}+\dfrac{3}{5x}=0\Rightarrow5x+9=0\Leftrightarrow x=-\dfrac{9}{5}\)
a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)
=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)
=>\(6x-\dfrac{39}{4}=1\)
=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)
=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)
b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)
=>\(2x-6=3x+6-x+1\)
=>2x-6=2x+7
=>-6=7(vô lý)
c: \(x\left(x+3\right)+x\left(x-2\right)=2x\left(x-1\right)\)
=>\(x^2+3x+x^2-2x=2x^2-2x\)
=>3x-2x=-2x
=>3x=0
=>x=0
d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)
=>\(3x^2-3x-2x-4-2x=x^2-x\)
=>\(3x^2-7x-4-x^2+x=0\)
=>\(2x^2-6x-4=0\)
=>\(x^2-3x-2=0\)
=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)
\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)
\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)
\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)
\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)
\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)
\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)
a) Đặt A(x)=0
\(\Leftrightarrow-4x-5=0\)
\(\Leftrightarrow-4x=5\)
hay \(x=-\dfrac{5}{4}\)
b) Đặt B(x)=0
\(\Leftrightarrow3\left(2x-1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow6x-3-2x-2=0\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
dễ
ai đi qua tick cho mình nha
ai tick thì may mắn trọn đời
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