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a) \(3\dfrac{4}{5}:\dfrac{8}{5}=0,25:x\)
\(\Rightarrow\dfrac{19}{5}.\dfrac{5}{8}=\dfrac{x}{4}\)
\(\Rightarrow\dfrac{x}{4}=2\Rightarrow x=8\)
b) \(2x+\dfrac{3}{24}=3x-\dfrac{1}{32}\)
\(\Rightarrow x=\dfrac{1}{8}+\dfrac{1}{32}=\dfrac{5}{32}\)
c) \(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\)
\(\Rightarrow221x-34=152x+380\)
\(\Rightarrow69x=414\Rightarrow x=6\)
Bài 2:
a) 11. x = -55
x = -55 : 11
x = -5
Vậy x = -5
b) -3 . x = -12
x = -12 : (-3)
x = 4
Vậy x = 4
c) 3x - 12 = -48
3x = -48 + 12
3x = -36
x = -36 : 3
x = -12
Vậy x = -12
d) 10 + 23 . (2x - 10) = -36
23 . (2x - 10) = -36 - 10
23 . (2x - 10) = -46
2x - 10 = -46 : 23
2x - 10 = -2
2x = -2 + 10
2x = 8
x = 8: 2
x = 4
Vậy x = 4
f) | 2x - 1| + 3 = 8
|2x - 1| = 8 - 3
|2x - 1| = 5
\(\Rightarrow\left[\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy x = 3 hoặc x = -2
g) |2x2 - 3| - 4 = 11
|2x2- 3| = 11 + 4
|2x2 - 3| = 15
\(\Rightarrow\left[\begin{matrix}2x^2-3=15\\2x^2-3=-15\end{matrix}\right.\Rightarrow\left[\begin{matrix}2x^2=18\\2x^2=-12\end{matrix}\right.\Rightarrow\left[\begin{matrix}x^2=9\\x^2=-6\end{matrix}\right.\Rightarrow\left[\begin{matrix}x^2=3^2\\\left(loại\right)\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=3\\\end{matrix}\right.\)
Vậy x = 3.
Bài 1:
\((-48) . 72 +36 . (-304) \)
\(=\left[\left(-48\right).\left(36.2\right)\right]+36.\left(-340\right)\)
\(=\left[\left(-48\right).2\right].36+36.\left(-340\right)\)
\(=\left(-96\right).36+36.\left(-304\right)\)
\(=36.\left[\left(-96\right)+\left(-304\right)\right]\)
\(=36.\left(-400\right)\)
\(=-14400\)
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Tìm x
a, 5x - 10 . (-7) = 15 . (-3)
5x + 70 = -45
5x = -45 - 70
5x = -115
x = -115 : 5
x = -23
Vậy x = -23.
b, 2x + 12 = -8
2x = -8 - 12
2x = -20
x = -20 : 2
x = -10
Vậy x = -10.
c, 3x - 5 = -20
3x = -20 + 5
3x = -15
x = -15 : 3
x = -5
Vậy x = -5.
d, 2x - (-12) = -8 . 23
2x + 12 = -184
2x = -184 - 12
2x = -196
x = -196 : 2
x = -98
Vậy x = -98.
e, x2 = 49
x2 = 72 = (-7)2
=> x = 7 hoặc x = -7.
Vậy x thuộc {-7 ; 7}
i, (x - 2)2 + 1 = 65
(x - 2)2 = 65 - 1
(x - 2)2 = 64
(x - 2)2 = 82 = (-8)2
=> x - 2 = 8 hoặc x - 2 = -8
=> x = 10 hoăc x = -6
Vậy x thuộc {-6 ; 10}.
g, 2 (x - 3)2 - 5 = 5 . (-3)2
2 (x - 3)2 - 5 = 5 . 9
2 (x - 3)2 - 5 = 45
2 (x - 3)2 = 45 + 5
2 (x - 3)2 = 50
(x - 3)2 = 50 : 2
(x - 3)2 = 25
(x - 3)2 = 52 = (-5)2
=> x - 3 = 5 hoặc x - 3 = -5
=> x = 8 hoặc x = -2
Vậy x thuộc {-2 ; 8}.
a) 5x - 10.(-7)=15.(-3) b) 2x+12=-8 c) 3x-5=-20 d)2x-(-12)=-8.23 e)\(^{x^2}\)=49 i)\(\left(x-2\right)^2+1\)=65
5x+70=-45 2x=-8-12 3x=-20+5 2x+12=-184 x=7 \(\left(x-2\right)^2\) =64
5x=-45-70 2x=-20 3x=-15 2x=-184-12 X-2 =8
5x=-115 x=-20:2 x=-15:3 2x=-186 x =10
x=-23 x=-10 x=-5 x=-93
g)\(2.\left(x-3\right)^2-5=5.\left(-3\right)^2\)
\(2\left(x-3\right)^2-5=30\)
\(2.\left(x-3\right)^2=35\)
\(2.\left(x^2-2x.3+3^2\right)=35\)
\(2x^2-24x+18=35\)
\(2x^2-24x=17\)
\(2x\left(x-12\right)=17\)
\(\orbr{\begin{cases}2x=17\\x-12=17\end{cases}}\orbr{\begin{cases}x=\frac{17}{2}\\x=29\end{cases}}\)
X
2: =>2x^2-8x+4=x^2-4x+4 và x>=2
=>x^2-4x=0 và x>=2
=>x=4
3: \(\sqrt{x^2+x-12}=8-x\)
=>x<=8 và x^2+x-12=x^2-16x+64
=>x<=8 và x-12=-16x+64
=>17x=76 và x<=8
=>x=76/17
4: \(\sqrt{x^2-3x-2}=\sqrt{x-3}\)
=>x^2-3x-2=x-3 và x>=3
=>x^2-4x+1=0 và x>=3
=>\(x=2+\sqrt{3}\)
6:
=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=-2\)
=>\(\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|=-2\)
=>\(\left|\sqrt{x-1}-1\right|=\sqrt{x-1}+1+2=\sqrt{x-1}+3\)
=>1-căn x-1=căn x-1+3 hoặc căn x-1-1=căn x-1+3(loại)
=>-2*căn x-1=2
=>căn x-1=-1(loại)
=>PTVN
1) ĐK: \(x\ge\dfrac{5}{2}\)
pt <=> \(x-4=\sqrt{2x-5}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left(x-4\right)^2=2x-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-8x+16=2x-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-10x+21=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left(x-3\right)\left(x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left[{}\begin{matrix}x=3\left(l\right)\\x=7\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy, pt có nghiệm duy nhất là x=7
2) ĐK: \(2x^2-8x+4\ge0\)
pt <=> \(\left\{{}\begin{matrix}x\ge2\\2x^2-8x+4=x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\left(x-4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=0\left(l\right)\\x=4\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy, pt có nghiệm duy nhất là x=4
3) ĐK: \(x\ge3\)
pt <=> \(\left\{{}\begin{matrix}x\le8\\x^2+x-12=x^2-16x+64\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\17x=76\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x=\dfrac{76}{17}\left(n\right)\end{matrix}\right.\)
Vậy, pt có nghiệm duy nhất là \(x=\dfrac{76}{17}\)\(\)
a) \(3x-10=2x+13\Leftrightarrow x=13+10=23\)
b) \(x+12=-5-x\Leftrightarrow2x=-17\Leftrightarrow x=-\frac{17}{2}\)
c) \(x+5=10-x\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\)
d) \(6x+23=2x-12\Leftrightarrow4x=-35\Leftrightarrow x=-\frac{35}{4}\)
e) \(12-x=x+1\Leftrightarrow2x=11\Leftrightarrow x=\frac{11}{2}\)
f) \(14+4x=3x+20\Leftrightarrow x=20-14=6\)
g) \(2\left(x-1\right)+3\left(x-2\right)=x-4\Leftrightarrow2x-2+3x-6=x-4\)
\(\Leftrightarrow5x-8=x-4\Leftrightarrow4x=4\Leftrightarrow x=1\)
h ) \(3\left(4-x\right)-2\left(x-1\right)=x+20\Leftrightarrow12-3x-2x+2=x+20\)
\(\Leftrightarrow14-5x=x+20\Leftrightarrow6x=-6\Leftrightarrow x=-1\)
i ) \(4\left(2x+7\right)-3\left(3x-2\right)=24\Leftrightarrow8x+28-9x+6=24\)
\(\Leftrightarrow34-x=24\Leftrightarrow x=34-24=10\)
k) \(3\left(x-2\right)+2x=10\Leftrightarrow3x-6+2x=5x-6=10\)
\(\Leftrightarrow5x=10+6=16\Leftrightarrow x=\frac{16}{5}\)
Tự KL cho các phần
2. 3x = 162
3x = 162 : 2
3x = 81
3x = 34
=> x = 4
2x + 23 = 23 . 32
2x + 8 = 72
2x = 72 - 8
2x = 64
2x = 26
=> x = 6
a) 2.3x = 162
3x = 81 = 34
=> x = 4
b) 2x + 23 = 23.32
2x = 64 = 26
=> x = 6
c, d bn lm tương tự như phần a nha
e) (1+2x)3 = 729 = 93
=> 1 + 2x = 9
2x = 8
x = 4
f) 2x + 2x+3 = 144
2x.(1+23) = 144
2x.9 = 144
2x = 16 = 24
=> x = 4