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a) \(\left(x-1\right)^3=8=2^3\)
\(x-1=2\)
\(x=2+1=3\)
b) \(7^{2x-6}=49=7^2\)
\(2x-6=2\)
\(2x=6+2=8\)
\(x=8:2=4\)
c) \(\left(2x-14\right)^7=128=2^7\)
\(2x-14=2\)
\(2x=14+2=16\)
\(x=16:2=8\)
d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)
\(x=5\)
e) \(3\cdot\left(x+2\right):7\cdot4=120\)
\(x+2=120:3\cdot7:4\)
\(x+2=70\)
\(x=70-2=68\)
Lời giải:
a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$
$\Rightarrow x=3$
b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$
$\Rightarrow 2x=8$
$\Rightarrow x=4$
c. $(2x-14)^7=128=2^7$
$\Rightarrow 2x-14=2$
$\Rightarrow 2x=16$
$\Rightarrow x=18$
d.
$x^4.x^5=5^3.5^6$
$x^9=5^9$
$\Rightarrow x=5$
e.
$3(x+2):7=120:4=30$
$3(x+2)=30.7=210$
$x+2=210:3=70$
$x=70-2=68$
a ) Ta có : 4(x - 5) - 3(x + 7) = -19
<=> 4x - 20 - 3x - 21 = -19
=> x - 41 = -19
=> x = -19 + 41
=> x = 22
b) Ta có " 7(x - 3) - 5(3 - x) = 11x - 5
<=> 7x - 21 - 15 + 5x = 11x - 5
<=> 12x - 36 = 11x - 5
=> 12x - 11x = -5 + 36
=> x = 31
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
2:
a: =>2/3:x=1,4-2,4=-1
=>x=-2/3
b: =>x/5=25/30-19/30=6/30=1/5
=>x=1
3:
Số học sinh giỏi là 40*1/4=10 bạn
Số học sinh khá là 30*3/5=18 bạn
Số học sinh TB là 30-18=12 bạn
\(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\\ =>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\\ =>2x-\dfrac{1}{2}=\dfrac{10}{12}-\dfrac{15}{12}\\ =>2x-\dfrac{1}{2}=-\dfrac{5}{12}\\ =>2x=-\dfrac{5}{12}+\dfrac{1}{2}\\ =>2x=-\dfrac{5}{12}+\dfrac{6}{12}\\ =>2x=\dfrac{1}{12}\\ =>x=\dfrac{1}{12}:2\\ =>x=\dfrac{1}{12}\cdot\dfrac{1}{2}\\ =>x=\dfrac{1}{24}\)
__
\(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{12}{8}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{7}{8}\\ =>x=\dfrac{7}{8}-\dfrac{1}{4}\\ =>x=\dfrac{7}{8}-\dfrac{2}{8}\\ =>x=\dfrac{5}{8}\)
__
\(\dfrac{x}{3}=\dfrac{12}{x}\\ =>x^2=3\cdot12\\ =>x^2=36\\ =>x^2=6^2\\ =>x=\pm6\)
Tìm x:
a) \(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\)
\(=>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\)
\(=>2x-\dfrac{1}{2}=\dfrac{-5}{12}\)
\(=>2x=\dfrac{-5}{12}+\dfrac{1}{2}\)
\(=>2x=\dfrac{1}{12}\)
\(=>x=\dfrac{1}{12}:2\)
\(=>x=\dfrac{1}{24}\)
b) \(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{7}{8}\)
\(=>x=\dfrac{7}{8}-\dfrac{1}{4}\)
\(=>x=\dfrac{5}{8}\)
c) \(\dfrac{x}{3}=\dfrac{12}{x}\)
Ta có: \(x.x=3.12\)
\(\Rightarrow x^2=36\)
Vậy x = 6 hoặc x = -6
Chúc bạn học tốt
\(a.\Leftrightarrow2|x-6|-|x-6|-2=0\)
\(\Leftrightarrow|x-6|-2=0\)
\(\Leftrightarrow|x-6|=2\)
\(+x-6=2\)
\(\Leftrightarrow x=8\)
\(+x-6=-2\)
\(\Leftrightarrow x=4\)
v...
\(b.\Leftrightarrow-4\left(5-x\right)-7\left(5-x\right)+10\left(5-x\right)=-3\)
\(\Leftrightarrow\left(5-x\right)\left(10-4-7\right)=-3\)
\(\Leftrightarrow-1.\left(5-x\right)=-3\)
\(\Leftrightarrow5-x=3\)
\(\Leftrightarrow x=2\)
v...
a) \(\left|\frac{4}{7}-x\right|+\frac{2}{5}=0\)
=> \(\left|\frac{4}{7}-x\right|=-\frac{2}{5}\), vô lí vì \(\left|\frac{4}{7}-x\right|\ge0\)
Vậy không tồn tại giá trị của x thỏa mãn đề bài
b) \(6-\left|\frac{1}{4}x+\frac{2}{5}\right|=0\)
=> \(\left|\frac{1}{4}x+\frac{2}{5}\right|=6-0=6\)
=> \(\left[\begin{array}{nghiempt}\frac{1}{4}x+\frac{2}{5}=6\\\frac{1}{4}x+\frac{2}{5}=-6\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}\frac{1}{4}x=\frac{28}{5}\\\frac{1}{4}x=-\frac{32}{5}\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=\frac{112}{5}\\x=-\frac{128}{5}\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=\frac{112}{5}\\x=-\frac{128}{5}\end{array}\right.\)
c) \(\left|x-\frac{1}{3}\right|+\left|2-\frac{4}{5}\right|=0\)
=> \(\left|x-\frac{1}{3}\right|+\left|\frac{6}{5}\right|=0\)
=> \(\left|x-\frac{1}{3}\right|+\frac{6}{5}=0\)
=> \(\left|x-\frac{1}{3}\right|=-\frac{6}{5}\), vô lí vì \(\left|x-\frac{1}{3}\right|\ge0\)
Vậy không tồn tại giá trị của x thỏa mãn đề bài
a: \(\Leftrightarrow2^x=1024\cdot3+1024\cdot7776+7776\cdot5\)
\(\Leftrightarrow2^x=8004576\)
hay \(x\in\varnothing\)
b: \(\Leftrightarrow x\left(x+3\right)^{100}-\left(x+3\right)^{100}=0\)
\(\Leftrightarrow\left(x+3\right)^{100}\left(x-1\right)=0\)
=>x=-3 hoặc x=1