tìm x
a, 3/4 + -1/2x = 1
b, 1/6 :x -1/3 = 1/2
c,(x+1/5)2=9
d,22/9-(x+1/2)2=7/3
e, 2|x|+1/2=2
f,|x+1/2|-1/6=1
giải giúp mình vs mk đang cần gấp
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f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
a)x7+x5+1=x7+x6-x6+2x5-x5+x4-x4+x3-x3+x2-x2+1
=x7-x6+x5-x3+x2+x6-x5+x4-x2+x+x5-x4+x3-x+1
=x2(x5-x4+x3-x+1)+x(x5-x4+x3-x+1)+1(x5-x4+x3-x+1)
=(x2+x+1)(x5-x4+x3-x+1)
b)4x4-32x2+1=4x4+12x3+2x2-12x3-36x2-6x+2x2+6x+1
=2x2(2x2+6x+1)-6x(2x2+6x+1)+1(2x2+6x+1)
=(2x2-6x+1)(2x2+6x+1)
c)x6+27=(x2+3)(x2-3x+3)(x2+3x+3)
d)3(x4+x2+1)-(x2+x+1)
=3x4-3x3+2x2+3x3-3x2+2x+3x2-3x+2
=x2(3x2-3x+2)+x(3x2-3x+2)+1(3x2-3x+2)
=(x2+x+1)(3x2-3x+2)
e)bạn tự làm nhé
\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
Câu a) xem lại đề giùm nhé em
b) \(\left(x-1\right)^3=9^3\)
\(x-1=9\)
\(x=10\)
Vậy \(x=10\)
c) \(\left(x-1\right)^2=25\)
\(x-1=5\) hoặc \(x-1=-5\)
* \(x-1=5\)
\(x=6\)
* \(x-1=-5\)
\(x=-4\)
Vậy \(x=-4\); \(x=6\)
d) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy \(x=2\)
e) Sửa đề: \(\left(2x+4\right)^3=64\)
\(\left(2x+4\right)^3=4^3\)
\(2x+4=4\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)
a: x/3-1/6=1/5
=>x/3=11/30
hay x=11/90
b: =>1/2x=2
hay x=4
c: =>2/3:x=-7-1/3=-22/3
=>x=-1/11
1. \(3-|2x+1|=-5\)
\(\Rightarrow|2x+1|=8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)
2.\(12+|3-x|=9\)
\(\Rightarrow|3-x|=-3\)
Mà \(|3-x|\ge0\forall x\)
\(\Rightarrow\)Vô lí
Vậy không có x
3.\(|x+9|=12+\left(-9\right)+2\)
\(\Rightarrow|x+9|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
4.\(5x-16=40+x\)
\(\Rightarrow5x-x=40+16\)
\(\Rightarrow4x=56\)
\(\Rightarrow x=14\)
Vậy \(x=14\)
5.\(5x-7=-21-2x\)
\(\Rightarrow5x+2x=-21+7\)
\(\Rightarrow7x=-14\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
6.\(\left(2x-1\right)\left(y-2\right)=12\)
Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)
\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng : (em tự xét bảng nhé)
a. 5 - 3(x + 4) = -1
⇔ 5 - 3x - 12 = -1
⇔ 3x = -1 - 5 + 12
⇔ 3x = 6
⇔ x = 2
\(d,2x^2-3=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
\(e,x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
1:
a: =>(|x|+4)(|x|-1)=0
=>|x|-1=0
=>x=1; x=-1
b: =>x^2-4>=0
=>x>=2 hoặc x<=-2
d: =>|2x+5|=2x-5
=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0
=>x=0(loại)
Lời giải:
a. ĐKXĐ: $x\geq -9$
PT $\Leftrightarrow x+9=7^2=49$
$\Leftrightarrow x=40$ (tm)
b. ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$
$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$
$\Leftrgihtarrow 3\sqrt{2x+3}=15$
$\Leftrightarrow \sqrt{2x+3}=5$
$\Leftrightarrow 2x+3=25$
$\Leftrightarrow x=11$ (tm)
c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{2}{3}\)
d. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)
\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)
\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)
\(\Leftrightarrow -1=9\) (vô lý)
Vậy pt vô nghiệm.
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)