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a)Viết dưới dạng phân số rồi sử dụng tích chéo ý
b)\(\frac{-1}{7}.2^3-2x:1\frac{4}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-2x:\frac{7}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-\frac{6x}{7}=-2^{x-1}\)
\(\Rightarrow\frac{-8-6x}{7}=\frac{2^{x-1}}{-1}\)
\(\Rightarrow-1\left(-8-6x\right)=7.2^{x-1}\)
\(\Rightarrow6x+8=7.2^{x-1}\)
.........
a) \(\left(x-\frac{1}{2}\right)=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=0+\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}.\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow x-2=\pm1.\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1+2\\x=\left(-1\right)+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{3;1\right\}.\)
c) \(\left(x-2\right)^3=-8\)
\(\Rightarrow\left(x-2\right)^3=\left(-2\right)^3\)
\(\Rightarrow x-2=-2\)
\(\Rightarrow x=\left(-2\right)+2\)
\(\Rightarrow x=0\)
Vậy \(x=0.\)
Chúc bạn học tốt!
a,x=1:2 b,x=3 c,x=-4 d,x=2 e,x=2:3
mình làm thế là tắt còn tự nghĩ cách trình bày nhé!
Bài 1 :
a. \(\left|x-\frac{1}{3}\right|< \frac{5}{2}\)
TH1 : nếu \(\left|x-\frac{1}{3}\right|>0\)
\(x-\frac{1}{3}< \frac{5}{3}\)
\(x< 2\)
TH2 : nếu \(\left|x-\frac{1}{3}\right|< 0\)
\(\frac{1}{3}-x< \frac{5}{3}\)
\(x>-\frac{4}{3}\)
Bài 2 :
a. \(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2-1=0\)
\(\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\left(x-3\right)\left(x-1\right)=0\)
\(\left[\begin{array}{nghiempt}x-3=0\\x-1=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\)
a) ( x - 1/5 )2 = 0
<=> x - 1/5 = 0
<=> x = 1/5
b) ( x - 2 )2 = 1
<=> ( x - 2 )2 = ( ±1 )2
<=> x - 2 = 1 hoặc x - 2 = -1
<=> x = 3 hoặc x = 1
c) ( 2x - 1 )3 = -8
<=> ( 2x - 1 )3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -1
<=> x = -1/2
d) ( x4 )2 = x12/x5
<=> x8 = x7
<=> x8 - x7 = 0
<=> x7( x - 1 ) = 0
<=> x7 = 0 hoặc x - 1 = 0
<=> x = 0 hoặc x = 1
e) x10 = 25x8
<=> x10 - 25x8 = 0
<=> x8( x2 - 25 ) = 0
<=> x8 = 0 hoặc x2 - 25 = 0
<=> x = 0 hoặc x = ±5
f) ( 2x + 3 )2 = 9/121
<=> ( 2x + 3 )2 = ( ±3/11 )2
<=> 2x + 3 = 3/11 hoặc 2x + 3 = -3/11
<=> x = -15/11 hoặc x = -18/11
a) \(\left(x-\frac{1}{5}\right)^2=0\Leftrightarrow x-\frac{1}{5}=0\Leftrightarrow x=\frac{1}{5}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
c) \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3+8=0\)
\(\Leftrightarrow\left(2x-1+8\right)\left[\left(2x-1\right)^2-8\left(2x-1\right)+64\right]=0\)
\(\Leftrightarrow2x+7=0\)
\(\Leftrightarrow x=\frac{-7}{2}\)
d) ĐKXĐ : \(x\ne0\)
\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)
\(\Leftrightarrow x^8=x^7\)
\(\Leftrightarrow x^8-x^7=0\)
\(\Leftrightarrow x^7\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=1\left(tm\right)\end{cases}\Leftrightarrow x=1}\)
e) ĐKXĐ : x khác 0
\(x^{10}=25x^8\)
\(\Leftrightarrow x^2=25\Leftrightarrow x=5\)
f) \(\left(2x+3\right)^2=\frac{9}{121}\)
\(\Leftrightarrow\left(2x+3+\frac{3}{11}\right)\left(2x+3-\frac{3}{11}\right)=0\)
\(\Leftrightarrow\left(2x+\frac{36}{11}\right)\left(2x+\frac{30}{11}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-18}{11}\\x=-\frac{15}{11}\end{cases}}\)
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
1) a) \(x^2=2x\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) vậy \(x=0;x=2\)
b) \(x^3=x\Leftrightarrow x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\) \(\Leftrightarrow x\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\) vậy \(x=0;x=-1;x=1\)
\(x^2=2x\Rightarrow x^2-2x=0\Rightarrow x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\Rightarrow x=2\end{matrix}\right.\)
\(x^3=x\Rightarrow x^3-x=0\Rightarrow x\left(x^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\Rightarrow x^2=1\Rightarrow x=\pm1\end{matrix}\right.\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\left(\dfrac{1}{25}-1\right)...\left(\dfrac{1}{121}-1\right)\)
\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}.\dfrac{-24}{25}...\dfrac{-120}{121}\)
\(A=\dfrac{3.8.15.24....120}{4.9.16.25...121}\)
\(A=\dfrac{1.3.2.4.3.5.4.6....10.12}{2.2.3.3.4.4.5.5....11.11}\)
\(A=\dfrac{1.2.4....10}{2.3.4.5...11}.\dfrac{3.4.5....12}{2.3.4.5....11}\)
\(A=\dfrac{1}{11}.6=\dfrac{6}{11}\)
3) Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{8^{2017}+1}{8^{2018}+1}< 1\)
\(B< \dfrac{8^{2017}+1+8}{8^{2018}+1+8}\)
\(B< \dfrac{8^{2017}+8}{8^{2018}+8}\)
\(B< \dfrac{8\left(8^{2016}+1\right)}{8\left(8^{2017}+1\right)}\)
\(B< \dfrac{8^{2016}+1}{8^{2017}+1}=A\)
\(B< A\)
Bài 1 :
a/ \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
Vậy....
b/ \(x^2-10x+9=0\)
\(\Leftrightarrow x^2-9x-x+9=0\)
\(\Leftrightarrow x\left(x-9\right)-\left(x-9\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy...
c/ \(x^2+9x+8=0\)
\(\Leftrightarrow x^2+8x+x+8=0\)
\(\Leftrightarrow\left(x+8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-1\end{matrix}\right.\)
Vậy ...
d/ \(x^2-11x+10=0\)
\(\Leftrightarrow x^2-11x+10=0\)
\(\Leftrightarrow x^2-x-10x+10=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=10\end{matrix}\right.\)
Vậy...
Bài 2 :
Ta có :
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Leftrightarrow6x-3y=2x+2y\)
\(\Leftrightarrow6x-2x=2y+3y\)
\(\Leftrightarrow4x=5y\)
\(\Leftrightarrow\frac{x}{y}=\frac{5}{4}\)
Vậy....
Bài 3 : không hiểu đề lắm ???!!!!
Bài 4 :
Ta có :
\(\frac{x}{y^2}=2\Leftrightarrow x=2y^2\left(1\right)\)
Thay (1) ta có :
\(\frac{x}{y}=16\)
\(\Leftrightarrow\frac{2y^2}{y}=16\)
\(\Leftrightarrow2y=16\)
\(\Leftrightarrow y=8\Leftrightarrow x=128\)
Vậy...
a.
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
b.
\(\left(x-2\right)^2=1\)
\(x-2=\pm1\)
TH1:
\(x-2=1\)
\(x=1+2\)
\(x=3\)
TH2:
\(x-2=-1\)
\(x=-1+2\)
\(x=1\)
Vậy x = 3 hoặc x = 1
c.
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=-2+1\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d.
\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)
\(x+\frac{1}{2}=\pm\frac{1}{4}\)
TH1:
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(x=\frac{1}{4}-\frac{2}{4}\)
\(x=-\frac{1}{4}\)
TH2:
\(x+\frac{1}{2}=-\frac{1}{4}\)
\(x=-\frac{1}{4}-\frac{1}{2}\)
\(x=-\frac{1}{4}-\frac{2}{4}\)
\(x=-\frac{3}{4}\)
Vậy \(x=-\frac{1}{4}\) hoặc \(x=-\frac{3}{4}\)
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