Tìm x biết 3x+2/x-2=5/4
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1/4 - 5/2 x |3x - 1/5|=2/3 x |3x - 1/5|- 2/3
Tương đương với 1/4+2/3 = 2/3 x l3x - 1/5l + 5/2 x l3x-1/5l
11/12 = l3x - 1/5l x (2/3 + 5/2)
11/12 = l3x -1/5 l x 19/6
=> l3x - 1/5l = 11/12 : 19/6 = 11/38
Xét 2 trường hợp:
+ 3x - 1/5 = 11/38 => 3x = 11/38 + 1/5 = 93/190 => x = 93/190 : 3 = 31/190
+ 3x - 1/5 = -11/38 => 3x = -11/38 + 1/5 = -17/190 => x = -17/190 : 3 = -17/570
a. \(x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)
\(x^3-25x-\left(x^3+8\right)=17\)
\(x^3-25x-x^3-8=17\)
\(-25x=25\)
\(x=-1\)
c. \(6x^2-\left(6x^2-4x+15x-10\right)=7\)
\(6x^2-6x^2-11x+10=7\)
\(-11x=-3\)
\(x=\frac{3}{11}\)
1.
a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)
b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)
2.
a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)
b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)
a) (x-2)3+6(x+1)2-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0
\(\Rightarrow\)24x+10=0
\(\Rightarrow\)24x=-10
\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2
\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2
\(\Rightarrow\)x2-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2
\(\Rightarrow\)3x2-18x-22=3x2+2x+17
\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0
\(\Rightarrow\)-20x-39=0
\(\Rightarrow\)-20x=39
\(\Rightarrow\)x=\(-\dfrac{39}{20}\)
a) 3/2.|x - 5/3| - 4/5 = 4/3.|x - 5/3| + 1
<=> 3/2.|x - 5/3| = 4/3.|x - 5/3| + 1 + 4/5
<=> 3/2.|x - 5/3| = 9/5 + 4|x - 5/3|/3
<=> 3/2.|x - 5/3| - 4.|x - 5/3|/3 = 9/5
<=> |x - 5/3|/6 = 9/5
<=> |x - 5/3| = 9/5.6
<=> |x - 5/3| = 54/5
<=> x - 5/3 = 54/5 hoặc x - 5/3 = -54/5
x = 54/5 + 5/3 x = -54/5 - 5/3
x = 187/15 x = -137/15
b) 2.|3x + 1| = 1/3.|3x + 1| + 5
<=> 2.|3x + 1| - 1/3.|3x + 1| = 5
<=> 5/3.|3x + 1| = 5
<=> 5.|3x + 1| = 5.3
<=> 5.|3x + 1| = 15
<=> |3x + 1| = 15 : 5
<=> |3x + 1| = 3
3x + 1 = 3 hoặc 3x + 1 = -3
3x = 3 - 1 3x = -3 - 1
3x = 2 3x = -4
x = 2/3 x = -4/3
=> x = 2/3 hoặc x = -4/3
c) làm tương tự câu a) mình hơi lời
Làm câu c) cho
\(\frac{1}{4}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{4}+\frac{2}{3}=\frac{2}{3}\left|3x-\frac{1}{5}\right|+\frac{5}{2}\left|3x-\frac{1}{5}\right|\)
\(\Leftrightarrow\frac{3}{12}+\frac{8}{12}=\left|3x-\frac{1}{5}\right|\left(\frac{2}{3}+\frac{5}{2}\right)\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|\left(\frac{4}{6}+\frac{15}{6}\right)=\frac{11}{12}\)
\(\Leftrightarrow\frac{19}{6}\left|3x-\frac{1}{5}\right|=\frac{11}{12}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{12}.\frac{6}{19}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{38}\)
\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{5}=\frac{11}{38}\\3x-\frac{1}{5}=-\frac{11}{38}\end{cases}}\)
Giải tiếp nha
a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
$ a/ 12x(x – 5) – 3x(4x - 10) = 120$
`<=>12x^2-60x-12x^2+30x=120`
`<=>-30x=120`
`<=>x=-4`
Vậy `x=-4`
$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$
`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`
`<=>-6x^2+26x=112-6x^2-2x`
`<=>28x=112`
`<=>x=4`
Vậy `x=4`
$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$
`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`
`<=>-32x-18x^2=154+45x-18x^2`
`<=>77x=-154`
`<=>x=-2`
Vậy `x=-2`