1. Tìm x biết:
a) | 3/2.x + 1/2 | = | 4x - 1 |
b) | 5/4.x - 7/2 | - | 5/8.x + 3/5 | = 10
c) | 7/5.x + 2/3 | = | 4/5.x - 1/4 |
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\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a) X + 1/4 = 5/8
b) X - 3/5 = 1/10
c) X x 2/7 = 6/11
d) X : 3/2 = 1/4
đ) 7/4 - X = 5/7
e) 5/4 : x = 1/8
\(a,x+\dfrac{1}{4}=\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{2}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}\)
\(b,x-\dfrac{3}{5}=\dfrac{1}{10}\)
\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{6}{10}\)
\(\Leftrightarrow x=\dfrac{7}{10}\)
\(c,x\times\dfrac{2}{7}=\dfrac{6}{11}\)
\(\Leftrightarrow x=\dfrac{6}{11}:\dfrac{2}{7}\)
\(\Leftrightarrow x=\dfrac{6}{11}\times\dfrac{7}{2}\)
\(\Leftrightarrow x=\dfrac{42}{22}\)
\(\Leftrightarrow x=\dfrac{21}{11}\)
\(d,x:\dfrac{3}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{1}{4}\times\dfrac{3}{2}\)
\(\Leftrightarrow x=\dfrac{3}{8}\)
\(đ,\dfrac{7}{4}-x=\dfrac{5}{7}\)
\(\Leftrightarrow x=\dfrac{7}{4}-\dfrac{5}{7}\)
\(\Leftrightarrow x=\dfrac{49}{28}-\dfrac{20}{28}\)
\(\Leftrightarrow x=\dfrac{29}{28}\)
\(e,\dfrac{5}{4}:x=\dfrac{1}{8}\)
\(x=\dfrac{5}{4}:\dfrac{1}{8}\)
\(\Leftrightarrow x=\dfrac{5}{4}\times8\)
\(\Leftrightarrow x=\dfrac{40}{4}\)
\(\Leftrightarrow x=10\)
\(a,x=\dfrac{1}{2}-\dfrac{2}{5}\)
\(x=\dfrac{1}{10}\)
\(b,x+\dfrac{3}{7}=\dfrac{7}{10}\)
\(x=\dfrac{7}{10}-\dfrac{3}{7}\)
\(x=\dfrac{19}{70}\)
\(c,19-x=\dfrac{17}{20}\)
\(x=19-\dfrac{17}{20}\)
\(x=\dfrac{363}{20}\)
a) 2/3 x + 1/2 = 1/10
2/3 x = 1/10 - 1/2
2/3 x = -2/5
x = -2/5 : 2/3
x = -3/5
b) 2/3 x + 1/5 = 7/10
2/3 x = 7/10 - 1/5
2/3 x = 1/2
x = 1/2 : 2/3
x = 3/4
c) (3 4/5 - 2x) . 1 1/3 = 5
19/5 - 2x = 5 : 4/3
19/5 - 2x = 15/4
2x = 19/5 - 15/4
2x = 1/20
x = 1/20 : 2
x = 1/40
d) x/7 = 6/(-21)
x = 6.7/(-21)
x = -2
a) \(x+1^3=2^5-\left(-1^3\right)\)
\(\Rightarrow x+1=33\)
=> x = 32
b) \(3^7-x=1^4-\left(-3^5\right)\)
\(\Rightarrow2187-x=1+243=244\)
=> x = 1943
a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=-4x+1\end{cases}}\Rightarrow\orbr{\begin{cases}4x-\frac{3}{2}x-1=\frac{1}{2}\\-4x-\frac{3}{2}x+1=\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=\frac{3}{2}\\-\frac{11}{2}x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
phần b ở đề bài mình ghi sai, là bằng 0 chứ ko phải bằng 10