Giải các phương trình:
a) \(\left(x^2+2x+5\right)\left(x^2+4x\right)=0\)
b) \(\left(x^2-4x+4\right)\left(x^2-3x\right)=0\)
c) \(1,2x^3-x^2-0,2x=0\)
a.\(\left(x^2+2x+5\right)\left(x^2+4x\right)=0\)
Ta có: \(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\ge4>0;\forall x\)
\(\Rightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
b.\(\left(x^2-4x+4\right)\left(x^2-3x\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=3\end{matrix}\right.\)
c.\(1,2x^3-x^2-0,2x=0\)
\(\Leftrightarrow x\left(1,2x^2-x-0,2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-\dfrac{1}{6}\end{matrix}\right.\)
Bài 6: Rút gọn các biểu thức sau:
c) C = \(\left|x-7\right|+2x-3\)
Bài 7: Giải phương trình:
a) \(\left|0,5x-5\right|=2\)
b) \(\left|5x-2\right|=-3\)
c) \(\left|\dfrac{1}{4}x+3\right|=0\)
7:
a: =>0,5x-5=2 hoặc 0,5x-5=-2
=>0,5x=3 hoặc 0,5x=7
=>x=6 hoặc x=14
b: |5x-2|=-3
mà |5x-2|>=0
nên ptvn
c: =>1/4x+3=0
=>1/4x=-3
=>x=-12
giải các phương trình:
a)(x2+x+1)2-2x2-2x=5
b)\(\dfrac{1}{\left(x-1\right)\left(x-3\right)}\)+x2-4x+5=0
Giải phương trình:
a) \(5x^2-10x=4\left(x-1\right)\sqrt{x^2-2x+2}\)
b) \(\sqrt{2x^2+22x+29}-x-2=2\sqrt{2x+3}\)
c) \(x^3-7x^2+9x+12=\left(x-3\right)\left(x-2+5\sqrt{x-3}\right)\left(\sqrt{x-3}-1\right)\)
Mọi người giúp gấp với ạ.
Giải phương trình:
a) \(\sqrt{x^2+4}=\sqrt{2x+3}\)
b) \(\sqrt{x^2-6x+9}=2x-1\)
c) \(\sqrt{4x+12}=\sqrt{9x+17}-5\)
d) \(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
a: ĐKXĐ: x>=-3/2
\(\sqrt{x^2+4}=\sqrt{2x+3}\)
=>\(x^2+4=2x+3\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1(nhận)
b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))
=>\(\sqrt{\left(x-3\right)^2}=2x-1\)
=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>x=4/3(nhận) hoặc x=-2(loại)
c:
Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)
ĐKXĐ: \(x>=-3\)
\(\sqrt{4x+12}=\sqrt{9x+27}-5\)
=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)
=>\(-\sqrt{x+3}=-5\)
=>x+3=25
=>x=22(nhận)
d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)
=>\(4x^2-6x+1=4x^2-20x+25\)
=>\(-6x+20x=25-1\)
=>\(14x=24\)
=>x=12/7(nhận)
Giải các phương trình:
a) \(\left|2x\right|=x-6\)
b) \(\left|-3x\right|=x-8\)
c) \(\left|4x\right|=2x+12\)
d)\(\left|-5x\right|-16=3x\)
\(a,\left|2x\right|=x-6\)
\(\Leftrightarrow2x=x-6\)
\(\Leftrightarrow2x-x=-6\)
\(\Leftrightarrow x=-6\)
____________________
\(b,\left|-3x\right|=x-8\)
\(\Leftrightarrow3x=x-8\)
\(\Leftrightarrow3x-x=-8\)
\(\Leftrightarrow2x=-8\)
\(\Leftrightarrow x=-4\)
____________________
\(c,\left|4x\right|=2x+12\)
\(\Leftrightarrow4x=2x+12\)
\(\Leftrightarrow4x-2x=12\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
____________________
\(d,\left|-5x\right|-16=3x\)
\(\Leftrightarrow5x-16=3x\)
\(\Leftrightarrow5x-3x=16\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
Giải phương trình:
a, \(sin^2x+\left(1-\sqrt{3}\right)sinxcosx-\sqrt{3}cos^2x=0\).
b, \(3sin^2x-4sin\left(2x\right)+5cos^2x=2\).
a) \(sin^2x+\left(1-\sqrt[]{3}\right)sinxcosx-\sqrt[]{3}cos^2x=0\)
\(\Leftrightarrow tan^2x+\left(1-\sqrt[]{3}\right)tanx-\sqrt[]{3}=0\left(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt[]{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=tan\dfrac{3\pi}{4}\\tanx=tan\dfrac{\pi}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=tan\dfrac{3\pi}{4}+k\pi\\x=\dfrac{\pi}{3}+k\pi\end{matrix}\right.\left(k\in Z\right)\)
giải phương trình:
a,\(\left(\sqrt{1+x}+1\right)\left(\sqrt{1+x}+2x-5\right)=x\)
b, \(4\sqrt{6x+10}=4x^2+14x+11\)
a.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow\left(\sqrt{x+1}+1\right)\left(\sqrt{x+1}+2x-5\right)=x+1-1\)
\(\Leftrightarrow\left(\sqrt{x+1}+1\right)\left(\sqrt{x+1}+2x-5\right)=\left(\sqrt{x+1}+1\right)\left(\sqrt{x+1}-1\right)\)
\(\Leftrightarrow\sqrt{x+1}+2x-5=\sqrt{x+1}-1\)
\(\Leftrightarrow2x-5=-1\)
\(\Leftrightarrow x=2\)
b.
ĐKXĐ: \(x\ge-\dfrac{5}{3}\)
\(6x+10+4\sqrt{6x+10}+4=4x^2+20x+25\)
\(\Leftrightarrow\left(\sqrt{6x+10}+4\right)^2=\left(2x+5\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{6x+10}+4=2x+5\\\sqrt{6x+10}+4=-2x-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{6x+10}=2x+1\left(1\right)\\\sqrt{6x+10}=-2x-9< 0\left(loại\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow6x+10=4x^2+4x+1\) \(\left(x\ge-\dfrac{1}{2}\right)\)
\(\Leftrightarrow4x^2-2x-9=0\)
\(\Rightarrow x=\dfrac{1+\sqrt{37}}{4}\)
Giải các hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(3x+2\right)\left(2y-3\right)=6xy\\\left(4x+5\right)\left(y-5\right)=4xy\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2y-5x}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)
Giải phương trình:
a) \(x^2-\left(x+3\right)\left(3x+1\right)=9\).
b) \(x^3+4x+5=0\).
c) \(\left(x+14\right)^3-\left(x+12\right)^3=1352\).
d) \(x^3+\left(x-3\right)^3=\left(2x-3\right)^3\).
e) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=360\).
f) \(x^3+\left(x-2\right)\left(2x+1\right)=8\).
b) Ta có: \(x^3+4x+5=0\)
\(\Leftrightarrow x^3-x+5x+5=0\)
\(\Leftrightarrow x\left(x^2-1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+5\right)=0\)
mà \(x^2-x+5>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
a)x2-(x+3)(3x+1)=9
⇔(x-3)(x+3)-(x+3)(3x+1)=0
⇔x+3=0 hoặc 3x+1=0
1.x+3=0 ⇔x=-3
2.3x+1=0⇔x=-1/3
phương trình có 2 nghiệm x=-3 và x=-1/3
