2x+3 :8 = 128 :4
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\(a,128-3.\left(x+4\right)=23\\ \Rightarrow3.\left(x+4\right)=105\\ \Rightarrow x+4=35\\ \Rightarrow x=31\\ b,\left[\left(4x+28\right).3+55\right]:5=35\\ \Rightarrow\left(4x+28\right).3+55=175\\ \Rightarrow4x+28.3=120\\ \Rightarrow4x+28=60\\ \Rightarrow4x=32\\ \Rightarrow x=8.\)
c) \(\left(12x-4^3\right).8^3=4.8^4\)
\(12x-64=4.8^4:8^3\)
\(12x-64=32\)
\(12x=32+64\)
\(12x=96\)
\(x=\dfrac{96}{12}\)
\(x=8\)
d) \(720:\left[41-\left(2x-5\right)\right]:5=35\)
\(720:\left(41-2x+5\right):5=35\)
\(720:\left(46-2x\right)=35.5\)
\(720:\left(46-2x\right)=175\)
\(46-2x=720:175\)
\(46-2x=\dfrac{144}{35}\)
\(2x=46-\dfrac{144}{35}\)
\(2x=\dfrac{1466}{35}\)
\(x=\dfrac{1466}{35}:2\)
\(x=\dfrac{733}{35}\)
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a.
\(\Leftrightarrow\sqrt[3]{3x-5}=\left(2x-3\right)^3+2x-3-\left(3x-5\right)\)
Đặt \(\left\{{}\begin{matrix}2x-3=a\\\sqrt[3]{3x-5}=b\end{matrix}\right.\)
\(\Rightarrow b=a^3+a-b^3\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt[3]{3x-5}=2x-3\)
\(\Leftrightarrow3x-5=\left(2x-3\right)^3\)
\(\Leftrightarrow8x^3-36x^2+51x-22=0\)
\(\Leftrightarrow\left(x-2\right)\left(8x^2-20x+11\right)=0\)
\(\Leftrightarrow...\)
b.
\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x+3x-2-\sqrt[3]{81x-8}=0\)
\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x+\dfrac{\left(3x-2\right)^3-\left(81x-8\right)}{\left(3x-2\right)^2+\left(3x-2\right)\sqrt[3]{81x-8}+\sqrt[3]{\left(81x-8\right)^2}}=0\)
\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x+\dfrac{27\left(x^3-2x^2-\dfrac{5}{3}x\right)}{\left(3x-2\right)^2+\left(3x-2\right)\sqrt[3]{81x-8}+\sqrt[3]{\left(81x-8\right)^2}}=0\)
\(\Leftrightarrow\left(x^3-2x^2-\dfrac{5}{3}x\right)\left(1+\dfrac{27}{\left(3x-2\right)^2+\left(3x-2\right)\sqrt[3]{81x-8}+\sqrt[3]{\left(81x-8\right)^2}}\right)=0\)
\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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.
$3(x+2):7=120:4=30$
$3(x+2)=30.7=210$
$x+2=210:3=70$
$x=70-2=68$
![](https://rs.olm.vn/images/avt/0.png?1311)
128-3.(x+4) =23
3.(x+4) =128-23
3.(x+4) = 105
x+4 =105:3
x+4 =35
x = 35-4
x=31
![](https://rs.olm.vn/images/avt/0.png?1311)
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a,
128-3x-12=23
3x=128-12-23
3x=93
x=93:3
= 31
b,
(12x+84+55):5=35
12x+84+55=35.5
12x+84+55=175
12x=175-55-84
12x=36
x=36:12
x=3
\(2^{x+3}:8=128:4\)
\(\Rightarrow2^{x+3}:8=32\)
\(\Rightarrow2^{x+3}:2^3=2^5\)
\(\Rightarrow2^{x+3}=2^8\)
\(\Rightarrow x+3=8\)
\(\Rightarrow x=5\)
Vậy: \(x=5\)
\(2^{x+3}=256=2^8\)
\(\Leftrightarrow x+3=8\)
\(\Leftrightarrow x=5\)