(4-x)-2x(x-1)=X+20
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`#040911`
`a,`
`15 + 25 \div (2x - 1) = 20`
`\Rightarrow 25 \div (2x - 1) = 20 - 15`
`\Rightarrow 25 \div (2x - 1) = 5`
`\Rightarrow 2x - 1 = 25 \div 5`
`\Rightarrow 2x - 1 = 5`
`\Rightarrow 2x = 6`
`\Rightarrow x = 3`
Vây, `x = 3.`
`b,`
\(3^{x-1}+2\cdot3^x=21\)
`\Rightarrow 3^x \div 3 + 2. 3^x = 21`
`\Rightarrow 3^x . \frac{1}{3} + 2. 3^x = 21`
`\Rightarrow 3^x . (\frac{1}{3} + 2) = 21`
`\Rightarrow 3^x . \frac{7}{3} = 21`
`\Rightarrow 3^x = 21 \div \frac{7}{3}`
`\Rightarrow 3^x = 9`
`\Rightarrow 3^x = 3^2`
`\Rightarrow x = 2`
Vậy, `x = 2.`
`c,`
\(2^{x-3}+2^{x+1}=17\)
`\Rightarrow 2^x \div 2^3 + 2^x . 2 = 17`
`\Rightarrow 2^x . \frac{1}{8} + 2^x . 2 = 17`
`\Rightarrow 2^x . (\frac{1}{8} + 2) = 17`
`\Rightarrow 2^x . \frac{17}{8} = 17`
`\Rightarrow 2^x = 17 \div \frac{17}{8}`
`\Rightarrow 2^x = 8`
`\Rightarrow 2^x = 2^3`
`\Rightarrow x = 3`
Vậy, `x = 3`
`d,`
\(5^x-5^{x-1}=20\)
`\Rightarrow 5^x - 5^x \div 5 = 20`
`\Rightarrow 5^x - 5^x . \frac{1}{5} = 20`
`\Rightarrow 5^x . (1 - \frac{1}{5} = 20`
`\Rightarrow 5^x . \frac{4}{5} = 20`
`\Rightarrow 5^x = 20 \div \frac{4}{5}`
`\Rightarrow 5^x = 25`
`\Rightarrow 5^x = 5^2`
`\Rightarrow x = 2`
Vậy, `x = 2.`
\(a.25:\left(2x-1\right)=5\)
\(2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)
\(b.3^x:3+2.3^x=21\)\(\Leftrightarrow3^x.\dfrac{1}{3}+2.3^x=21\)
\(\Leftrightarrow3^x\left(\dfrac{1}{3}+2\right)=21\)
\(\Leftrightarrow3^x.\dfrac{7}{3}=21\)
\(\Leftrightarrow3^x=9\Leftrightarrow x=2\)
\(c.2^x:2^3+2^x.2=17\Leftrightarrow2^x.\dfrac{1}{8}+2^x.2=17\)
\(\Leftrightarrow2^x.\dfrac{17}{8}=17\Leftrightarrow2^x=8\Leftrightarrow x=3\)
\(d.5^x-5^x:5=20\Leftrightarrow5^x-5^x.\dfrac{1}{5}=20\)
\(\Leftrightarrow5^x\left(1-\dfrac{1}{5}\right)=20\Leftrightarrow5^x=20:\dfrac{4}{5}\Leftrightarrow5^x=25\Leftrightarrow x=2\)

mk làm lun nha
a, 2x^2-6x-3x-2x^2=26
-9x=26
x=-26/9
b,x^2+2.x.4+16-(x^2-1)=16
x^2+8x+16-x^2+1=16
8x=-1
x=-1/8
c,(2x)^2-2.2x.1+1-4(x^2-7^2)=0
4x^2-4x+1-4x^2+196=0
-4x=-197
x=197/4
d,x^2-5x-4x+20=0
-9x=-20
x=20/9
**** cho mk nha

a) 2x (x - 5) - x (3 + 2x) = 26
=> 2x2 - 10x - (3x - 2x2) = 26
=> 2x2 - 10x - 3x - 2x2 = 26
=> -13x = 26 => x = 26 : (-13) = -2
xin loi nhung hoi nhiu mik viet cau tra loi dc ko - Nguyễn Diệu Thảo

11: Ta có: \(\left(x+3\right)^3=125\)
\(\Leftrightarrow x+3=5\)
hay x=2
12: Ta có: \(\left(2x\right)^4=16\)
\(\Leftrightarrow x^4=1\)
hay \(x\in\left\{1;-1\right\}\)

\(2\left(x-1\right)+3\left(3x-2\right)=x-4\)
\(2x-2+9x-6=x-4\)
\(2x+9x-x-2-6=-4\)
\(10x-2-6=-4\)
\(10x-2=2\)
\(10x=4\)
\(x=\frac{2}{5}\)
Vậy \(x=\frac{2}{5}\)
\(3\left(4-x\right)-2\left(x-1\right)=x+20\)
\(12-3x-2x+2=x+20\)
\(12-5x+2=x+20\)
\(12-5x-x+2=20\)
\(12-6x+2=20\)
\(12-6x=18\)
\(6x=-6\)
\(x=-1\)
Vậy \(x=-1.\)
\(4\left(2x+7\right)-3\left(3x-2\right)=24\)
\(8x+28-9x+6=24\)
\(8x-9x+28+6=24\)
\(-x+34=24\)
\(-x=-10\)
\(x=10\)
Vậy \(x=10\)
\(3\left(x-2\right)+2x=10\)
\(3x-6+2x=10\)
\(3x+2x-6=10\)
\(5x=16\)
\(x=\frac{16}{5}\)
Vậy \(x=\frac{16}{5}\)
2(x-1)+3(3x-2)=x-4
=>2x-2=9x-6-x+4=0
=>10x-4=0
=>x=\(\frac{2}{5}\)

*\(3x-2\left(2x-1\right)=1\frac{1}{3}\)
\(3x-4x+2=\frac{4}{3}\)
\(-x+2=\frac{4}{3}\)
\(-x=\frac{4}{3}-2\)
\(-x=-\frac{2}{3}\)
\(\Rightarrow x=\frac{2}{3}\)
*\(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right).\left(x+4\right)=5.20\)
\(\left(x+4\right)^2=100\)
\(\left(x+4\right)^2=10^2\)
\(\Rightarrow x+4=10\)
\(\Rightarrow x=10-4=6\)

\(d,\left(x-20\right)\left(x^2+2x+4\right)-x^2\left(x+1\right)=-17\)
\(x^3+2x^2+4x-20x^2-40x-80-x^3-x^2=-17\)
\(-19x^2-36x-63=0\)
\(\Delta=\left(-36\right)^2-\left(-63.-19.4\right)=-3492< 0\)
pt vô nghiệm

Bài 1:
a. $(-20)+x=-30$
$x-20=-30$
$x=-30+20=-(30-20)=-10$
b.
$(-10)-x=-20$
$x=(-10)-(-20)=-10+20=20-10=10$
c. Đề sai. Bạn xem lại.
d.
$x+(-3)=-7$
$x=-7-(-3)=-7+3=-(7-3)=-4$
e.
$x-(-5)=-9$
$x=(-9)+(-5)=-14$
f.
$x(-11)=12$
$x=\frac{12}{-11}=\frac{-12}{11}$
h.
$2x-10=20$
$2x=20+10=30$
$x=30:2=15$
l.
$4x-8=-8$
$4x=-8+8=0$
$x=0:4=0$
k.
$-12-(-2)x=-8$
$(-2)x=-12-(-8)=-12+8=-(12-8)=-4$
$x=(-4):(-2)=2$
Bài 2:
a. $-20-(10-x)=-3$
$10-x=-20-(-3)=-20+3=-(20-3)=-17$
$x=10-(-17)=10+17=27$
b.
$14+(14-x)=-2$
$14-x=-2-14=-16$
$x=14-(-16)=14+16=30$
c.
$-15-(x-3)=-7$
$x-3=-15-(-7)=-15+7=-8$
x=-8+3=-5$
d.
$(x+4)+(-20)=-8$
$x+4=-8-(-20)=-8+20=12$
$x=12-4=8$
e.
$-2x-2=-4$
$-2x=-4+2=-2$
$x=(-2):(-2)=1$
f.
$-2x+4=-4$
$-2x=-4-4=-8$
$x=(-8):(-2)=4$
l.
$-12-(-2)x=-2-4=-6$
$(-2)x=-12-(-6)=-12+6=-6$
$x=(-6):(-2)=3$
\(\left(4-x\right)-2x\left(x-1\right)=x+20\)
=>\(4-x-2x^2+2x-x-20=0\)
=>\(-2x^2-16=0\)
=>\(x^2+8=0\)
mà \(x^2+8\ge8>0\forall x\)
nên x∈∅