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1: Ta có: \(20-2\left(x+4\right)=4\)
\(\Leftrightarrow2\left(x+4\right)=16\)
\(\Leftrightarrow x+4=8\)
hay x=4
5: Ta có: \(\left(x+1\right)^3=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
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a) | 5/4x -7/2| - | 5/8x + 3/5| = 0
|5/4x - 7/2| = | 5/8x + 3/5|
TH1: 5/4x - 7/2 = 5/8x + 3/5
=> 5/4x - 5/8x = 3/5 +7/2
5/8x = 41/10
x = 41/10:5/8
x = 164/25
TH2: 5/4x - 7/2 = -5/8x - 3/5
=> 5/4x + 5/8x = -3/5 +7/2
15/8x = 29/10
x = 29/10 : 15/8
x = 116/75
KL: x = 164/25 hoặc x = 116/75
các bài cn lại b lm tương tự nha! h lm dài lắm!
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\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)
\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)
\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)
\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)
\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)
\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)
\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)
\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)
\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)
\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)
\(< =>5x-35=32x-8< =>32x-5x=-35+8\)
\(< =>27x=-27< =>x=-1\)
\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)
\(< =>12=6x-3< =>6x=12+3\)
\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)
\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)
\(< =>8x-4=9x+3< =>9x-8x=-4-3\)
\(< =>9x-8x=-7< =>x=-7\)
\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)
\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)
\(< =>12x-7x=14-4< =>5x=10\)
\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)
\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)
\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)
\(< =>6x-4x=4+6< =>2x=10\)
\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)
\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)
\(< =>\left(x+1\right)\left(x+1\right)=3.3\)
\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)
\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)
\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)
\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)
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a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 )
<=> 2x - 3 - x + 5 = x + 7 - x - 2
<=> x = 3
b)(7x-5)-(6x+4)=(2x+3)-(2x+1)
<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1
<=> x = 11
c)(9x-3)-(8x+5)=(3x+2)
<=> 9x - 3 - 8x - 5 = 3x + 2
<=> -2x = 10
<=> x = -5
d)(x+7)-(2x+3)=(3x+5)-(2x+4)
<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4
<=> -2x = -3
<=> x = 3/2
ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠ
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(2x - 7) + 17 = 6
=> 2x - 7 = 6 - 17
=> 2x - 7 = -11
=> 2x = -11 + 7
=> 2x = -4
=> x = -4 : 2
=> x = -2
+) 12 -2(3 - 3x)= -2
=> 2(3 - 3x) = 12 + 2
=> 2(3 - 3x) = 14
=> 3 - 3x = 14 : 2
=> 3 - 3x = 7
=> 3x = 3 - 7
=> 3x = -4
=> x = -4/3
\(\left(x+1\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
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1) 4x-(2x-5)=21
4x-2x+5=21
2x+5=21
2x=21 -5
2x=16
x=16/2
x=8
=> 2x + 5 - 9x -21 = 6 - 6x + 8x =>-9x = 22 => x = -22/9
có đề = 2x + 5 - 9x -21 = 6 - 6x + 8x => -7x -2x = 6 + 16 => -9x = 22 => x = -22/9