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. 

$3(x+2):7=120:4=30$

$3(x+2)=30.7=210$

$x+2=210:3=70$

$x=70-2=68$

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mk chua hoc bai nay

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

a: \(\Leftrightarrow8x+16-5x^2-10x+4x^2-4x-8+2\left(x^2-4\right)=0\)

\(\Leftrightarrow-x^2-6x+8+2x^2-8=0\)

=>x^2-6x=0

=>x(x-6)=0

=>x=6 hoặc x=0

b: \(\Leftrightarrow24x^2+7x-6-4x^2-23x-28=10x^2+3x-1-33\)

\(\Leftrightarrow20x^2-16x-34-10x^2-3x+34=0\)

=>\(10x^2-19x=0\)

=>x(10x-19)=0

=>x=0 hoặc x=19/10

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

a)(x+3)³-x(3x+1)²+(2x+1)(4x²-2x+1-3x²)=54

\(\Rightarrow\)x³+9x²+27x+27-x(9x²+6x+1)+(2x+1)(x²-2x+1)=54

\(\Rightarrow\)x³+9x²+27x+27-9x³-6x²-x+2x³-4x²+2x+x²-2x+1=54

\(\Rightarrow\)-6x³+26x+28=54

\(\Rightarrow\)-6x³+26x=54-28

\(\Rightarrow\)-6x³+26x=26

\(\Rightarrow\)-6x³+26x-26=0

\(\Rightarrow\)-2(3x³+13x+14)

Đặt \(t=2x^2+3x-1\), pt trở thành:

\(t^2-7\left(t+4\right)+33=0\)

\(\Leftrightarrow t^2-7t+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{7+\sqrt{29}}{2}\\t=\dfrac{7-\sqrt{29}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2+3x-1=\dfrac{7+\sqrt{29}}{2}\\2x^2+3x-1=\dfrac{7-\sqrt{29}}{2}\end{matrix}\right.\)

Nghiệm xấu xuất sắc =]]

...

\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)

\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)

a) \(2x\left(x-\frac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0:2=0\\x=0+\frac{1}{7}=\frac{1}{7}\end{matrix}\right.\)

b) \(\frac{1}{2}x+\frac{3}{5}x=-\frac{33}{25}\)

\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{5}\right)=-\frac{33}{25}\)

\(\Rightarrow x\frac{11}{10}=-\frac{33}{25}\)

\(\Rightarrow x=\left(-\frac{33}{25}\right):\frac{11}{10}=-\frac{33}{25}.\frac{10}{11}=-\frac{6}{5}\)

c) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=0+\frac{4}{9}=\frac{4}{9}\\-\frac{3}{7}:x=0-\frac{1}{2}=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{9}:\frac{2}{3}=\frac{4}{9}.\frac{3}{2}=\frac{2}{3}\\x=\left(-\frac{3}{7}\right):\frac{-1}{2}=\left(-\frac{3}{7}\right).\left(-2\right)=\frac{6}{7}\end{matrix}\right.\)

a) \(2x\left(x-\frac{1}{7}\right)=0\)

⇒\(\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\)

Vậy \(x=0;x=\frac{1}{7}\)

b) \(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{25}\\ \left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{25}\\ \left(\frac{5}{10}+\frac{6}{10}\right)x=\frac{-33}{25}\\ \frac{11}{10}x=\frac{-33}{25}\\ x=\frac{-33}{25}:\frac{11}{10}\\ x=\frac{-33.10}{25.11}\\ x=\frac{-6}{5}\)

Vậy x = \(\frac{-6}{5}\)

c) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\\ \Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=\frac{4}{9}\\\frac{-3}{7}:x=\frac{-1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{4}{9}:\frac{2}{3}=\frac{4.3}{9.2}=\frac{2}{3}\\x=\frac{-3}{7}:\frac{-1}{2}=\frac{-3.2}{7.\left(-1\right)}=\frac{6}{7}\end{matrix}\right.\)