Tìm x,biết
a/x-2 = 2 1
3 3
b/3 1 = -0,25
4
c/ 50%.x=5 1 - 4 1
2 3
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\(a,\dfrac{3}{7}-x=\dfrac{1}{2}x-3\)
\(\Rightarrow-x-\dfrac{1}{2}x=-3-\dfrac{3}{7}\)
\(\Rightarrow-\dfrac{3}{2}x=-\dfrac{24}{7}\)
\(\Rightarrow x=-\dfrac{24}{7}:\left(-\dfrac{3}{2}\right)\)
\(\Rightarrow x=\dfrac{16}{7}\)
\(b,5x-\dfrac{2}{3}=\dfrac{5}{3}-2x\)
\(\Rightarrow5x+2x=\dfrac{5}{3}+\dfrac{2}{3}\)
\(\Rightarrow7x=\dfrac{7}{3}\)
\(\Rightarrow x=\dfrac{7}{3}:7\)
\(\Rightarrow x=\dfrac{1}{3}\)
#Toru
a: 3/7-x=1/2x-3
=>-3/2x=-3+3/7
=>-1/2x=-1+1/7=-6/7
=>1/2x=6/7
=>x=6/7*2=12/7
b: =>5x+2x=5/3+2/3
=>7x=7/3
=>x=1/3
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
a,C1
(5/6 + 5/8) x 2/3
= 35/24 x 2/3
=35/36
C2
(5/6 + 5/8) x 2/3
= 5/6x2/3 + 5/8x2/3
= 10/18 + 10/24
= 35/36
b,C1
7/5 x 3/4 - 1/2 x 3/4
= 21/20 - 3/8
= 27/40
C2
7/5 x 3/4 - 1/2 x 3/4
=3/4 x (7/5-1/2)
=3/4x9/10
=27/40
a, 5/6 x (1/8x2/3)
=5/6 x 1/12
=5/72
b,(7/5 - 1/2) x 3/4
=0,9 x 3/4
=3,06
c,(5/6+5/8) x 2/3
=70/48 x 2/3
=35/36
a, 5/6 x (1/8x2/3)
=5/6 x 1/12
=5/72
b,(7/5 - 1/2) x 3/4
=0,9 x 3/4
=3,06
c,(5/6+5/8) x 2/3
=70/48 x 2/3
=35/36
Bài 2:
a: \(\left(6x-39\right):7=3\)
\(\Leftrightarrow6x-39=21\)
hay x=10
a \(\dfrac{2}{3}x+\dfrac{1}{3}=\dfrac{1}{5}\\ \dfrac{2}{3}x=\dfrac{1}{5}-\dfrac{1}{3}\\ \dfrac{2}{3}x=\dfrac{-2}{15}\\ x=-\dfrac{2}{15}:\dfrac{2}{3}\\ x=-\dfrac{1}{5}\) b) \(\dfrac{4}{5}-\dfrac{5}{3}x=-2\\ \dfrac{5}{3}x=\dfrac{4}{5}+2\\ \dfrac{5}{3}x=\dfrac{14}{5}\\ x=\dfrac{14}{5}:\dfrac{5}{3}\\ x=\dfrac{42}{25}\)c) \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\\ \dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}\\ \dfrac{5}{3}:x=\dfrac{3}{10}\\ x=\dfrac{5}{3}:\dfrac{3}{10}\\ x=\dfrac{50}{9}\)d) \(\dfrac{5}{7}:x-3=-\dfrac{2}{7}\\ \dfrac{5}{7}:x=3-\dfrac{2}{7}\\ \dfrac{5}{7}:x=\dfrac{19}{7}\\ x=\dfrac{5}{7}:\dfrac{19}{7}\\ x=\dfrac{5}{19}\)
Lời giải:
a. $2x^2+3(x-1)(x+1)=5x(x+1)$
$\Leftrightarrow 2x^2+3x^2-3=5x^2+5x$
$\Leftrightarrow 5x^2-3=5x^2+5x$
$\Leftrightarrow 5x=-3$
$\Leftrightarrow x=\frac{-3}{5}$
b.
PT $\Leftrightarrow (-5x^2-2x+16)+4(x^2-x-2)=4-x^2$
$\Leftrightarrow -x^2-6x+8=4-x^2$
$\Leftrightarrow -6x+8=4$
$\Leftrightarrow -6x=-4$
$\Leftrightarrow x=\frac{2}{3}$
c.
PT $\Leftrightarrow 4(x^2+4x-5)-(x^2+7x+10)=3(x^2+x-2)$
$\Leftrightarrow 4x^2+16x-20-x^2-7x-10=3x^2+3x-6$
$\Leftrightarrow 3x^2+9x-30=3x^2+3x-6$
$\Leftrightarrow 6x=24$
$\Leftrightarrow x=4$
b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\dfrac{16}{8}=2\)
Do đó: x=5; y=5; z=17
\(a,\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)
Áp dụng t/c dtsbn:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+2y^2-3z^2}{4+18-48}=\dfrac{-650}{-26}=25\\ \Rightarrow\left\{{}\begin{matrix}x^2=100\\y^2=225\\z^2=400\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm10\\y=\pm15\\z=\pm20\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)\) có giá trị là hoán vị của \(\left(\pm10;\pm15;\pm20\right)\)
a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
a, \(\frac{x-2}{3}=2.\frac{1}{3}\Leftrightarrow\frac{x-2}{3}=\frac{2}{3}\)
\(\Leftrightarrow x-2=2\Leftrightarrow x=4\)
b, \(3.\frac{1}{4}=-0,25\Leftrightarrow\frac{3}{4}=-0,25\)đề sai rồi bn nhé
c, \(50\%x=5.\frac{1}{2}-4.\frac{1}{3}\Leftrightarrow50\%x=\frac{5}{2}-\frac{4}{3}\)
\(\Leftrightarrow50\%x=\frac{7}{6}\Leftrightarrow x=\frac{7}{3}\)