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a, 4x - 15 = ( -75 ) - x
<=> 4x +x = -75 + 15
<=> 5x = -60
<=> x = -12
b, 72 - 3x = 5x +8
<=> 72 - 8 = 5x + 3x
<=> 8x = 64
<=> x = 8
c, 3.|x-7| = 21
* Nếu x - 7 \(\ge\)0 <=> x \(\ge\)7 thì
PT <=> 3 .( x- 7 ) = 21
<=> x - 7 = 7
<=> x = 14 ( tm )
* Nếu x < 7 <=> x < 7 thì
PT <=> 3. [ - ( x- 7 ) ] = 21
<=> -x +7 = 7
<=> -x = 0
<=> x = 0 ( tm )
d, \(-7.|x+3|=-49\).
\(\Leftrightarrow|x+3|=7\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-10\end{cases}}}\)
e, c-12.(x-5)+7.(3-x)=5
<=> -12x + 60 + 21 - 7x = 5
<=> -19x + 81 = 5
<=> -19x = -76
<=> x = 4
2 x X + 68 = 126
2 x X = 126 - 68
2 x X = 58
x = 58 : 2
x = 29
Bài 1:
Ta có: \(\frac{497}{-499}=-\frac{497}{499}>-\frac{499}{499}=-1\left(1\right)\)
\(-\frac{2345}{2341}< -\frac{2341}{2341}=-1\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{497}{-499}>-\frac{2345}{2341}\)
Bài 2:
\(\frac{x+5}{2005}+\frac{x+6}{2004}=\frac{x+7}{2003}+3=0\)
\(\Rightarrow\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}+3=0\)
\(\Rightarrow\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=0\)
\(\Rightarrow\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)
\(\Rightarrow\left(x+2010\right)\times\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)
Vì \(\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\ne0\Rightarrow x+2010=0\)
\(\Rightarrow x=0-2010=-2010\)
Vậy x = -2010
a. \(\frac{7}{8}< \frac{x}{35}< \frac{15}{7}\)
\(\Rightarrow\frac{245}{280}< \frac{8x}{280}< \frac{600}{280}\)
\(\Rightarrow245< 8x< 600\)
\(\Rightarrow30< x< 75\)
\(\Rightarrow x\in\left\{31;32;33;...;72;73;74\right\}\)
b. \(\frac{21}{3}< \frac{x}{7}\le\frac{24}{2}\)
\(\Rightarrow\frac{294}{42}< \frac{6x}{42}\le\frac{504}{42}\)
\(\Rightarrow294< 6x\le504\)
\(\Rightarrow49< x\le84\)
\(\Rightarrow x\in\left\{50;51;52;...;82;83;84\right\}\)
\(\left(3x-5\right)^8=\frac{1}{125}\left(5-3x\right)^{11}\)
\(\Leftrightarrow-\left(3x-5\right)^8=-\frac{1}{125}\left(3x-5\right)^{11}\)
\(\Leftrightarrow-1=-\frac{1}{125}\left(3x-5\right)^3\)
\(\Leftrightarrow\frac{1}{125}\left(3x+5\right)^3=1\)
\(\Leftrightarrow\left(3x-5\right)^3=125\)
\(\Leftrightarrow3x-5=\sqrt[3]{125}\)
\(\Leftrightarrow3x-5=5\)
\(\Leftrightarrow3x=10\)
\(\Leftrightarrow x=\frac{10}{3}\)
Vậy phương trình đã cho có tập nghiệm \(S=\left\{\frac{10}{3}\right\}\)
a) 400 - 5x = 200
5x = 200
x = 40
b) 250 : x + 10 = 20
250 : x = 10
x = 25
c) 96 - 3 ( x + 8 ) = 42
3 ( x + 8 ) = 54
( x + 8 ) = 54 : 3
x + 8 = 18
x = 18 - 8
x = 10
d) 36 : ( x - 5 ) = 22
36 : ( x - 5 ) = 4
x - 5 = 36 : 4
x - 5 = 9
x = 9 + 5
x = 14
e) 15 x 5 ( x - 35 ) - 525 = 0
75 ( x - 35 ) - 525 = 0
75 ( x - 35 ) = 525
x - 35 = 7
x = 7 + 35
x = 42
f) [ 3 x ( 70 - x ) + 5 ] : 2 = 46
[ 3 x ( 70 - x ) + 5 ] = 92
3 x ( 70 - x ) = 87
70 - x = 87 : 3
70 - x = 29
x = 41
a)TH1 x>=3 \(\left|x-3\right|\)=x-3
pttt: x-3-2x=1 suy ra x=-4 <3 -> loại
TH2 x=< 3 pttt 3-x-2x=1 suy ra x =2/3 thỏa mãn
b) VT=\(\dfrac{4^{x+2}+4^{x+1}+4^x}{21}=\dfrac{4^x\left(4^2+4+1\right)}{21}=4^x\)
VP= \(\dfrac{3^{2x}+3^{2x+1}+3^{2x+3}}{31}=\dfrac{9^x\left(1+3+27\right)}{31}=9^x\)
vậy pt đã cho tương đương với 4^x=9^x \(\Leftrightarrow\left(\dfrac{4}{9}\right)\)^x =1 suy ra x =0
\(311-x+82=46\left(x-21\right)\)
<=> \(311+82-46+21=x+x\)
<=> \(2x=368\)
<=> \(x=184\)
\(-x+821+534=499+x-84\)
<=> \(-x-x=499-84-821-534\)
<=> \(-2x=-940\)
<=> \(x=470\)
\(-\left(x-3+85\right)=x+70-71-5\)
<=> \(-x+3-85=x-6\)
<=> \(-x-x=-6-3+85\)
<=> \(-2x=76\)
<=> \(x=-38\)
311−x+82=46(x−21)
<=> 311+82−46+21=�+�311+82−46+21=x+x
<=> 2�=3682x=368
<=> �=184x=184
−�+821+534=499+�−84−x+821+534=499+x−84
<=> −�−�=499−84−821−534−x−x=499−84−821−534
<=> −2�=−940−2x=−940
<=> �=470x=470
−(�−3+85)=�+70−71−5−(x−3+85)=x+70−71−5
<=> −�+3−85=�−6−x+3−85=x−6
<=> −�−�=−6−3+85−x−x=−6−3+85
<=> −2�=76−2x=76
<=> �=−38x=−38