1 :\(\frac{7}{20}\)

2 \(\frac{1}{4}\)

3 \(\frac{23}{2}\)

4 2187

5 64

6 x=16

7 x=\(\frac{-1}{243}\)

8 mϵ∅

cho mình hỏi cài này là j vậy

Đề 2

1) \(\frac{7}{20}.\)

2) \(\frac{1}{4}.\)

3) \(\frac{23}{2}.\)

4) \(2187.\)

5) \(64.\)

6) \(x=16.\)

7) \(x=\left(-\frac{1}{3}\right)^5\)

8) \(m\in\varnothing.\)

Chúc bạn học tốt!

1, \(\frac{3}{5}+\left(-\frac{1}{4}\right)=\frac{12}{20}+\left(-\frac{5}{20}\right)=\frac{7}{20}\)

2,\(\left(-\frac{5}{18}\right)\left(-\frac{9}{10}\right)=\frac{-5.\left(-9\right)}{18.10}=\frac{45}{180}=\frac{1}{4}\)

3,\(4\frac{3}{5}:\frac{2}{5}=\frac{23}{5}.\frac{5}{2}=\frac{23}{2}\)

a) vì | x + \(\frac{5}{3}\)| \(\ge\)0 nên A = | x + \(\frac{5}{3}\)| + 112 \(\ge\)112

dấu " = " xảy ra khi | x + \(\frac{5}{3}\)| = 0 hay x = \(\frac{-5}{3}\)

\(\Rightarrow\)GTNN của A là 112 khi | x + \(\frac{5}{3}\) | = 0 hay x = \(\frac{-5}{3}\)

b) B = | x - 2,7 | + | x + 8,5 |

B = | 2,7 - x | + | x + 8,5 | \(\ge\)| 2,7 - x + x + 8,5 | = 11,2

\(\Rightarrow\)GTNN của B là 11,2 khi ( 2,7 - x ) . ( x + 8,5 ) \(\ge\)0 hay -8,5 \(\le\)x \(\le\)2,7

c) C = \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|2x+\frac{1}{4}\right|\)

C = \(\left|x+\frac{1}{2}\right|+\left|-\frac{1}{3}-x\right|+\left|2x+\frac{1}{4}\right|\)\(\ge\)\(\left|x+\frac{1}{2}-\frac{1}{3}-x\right|+\left|2x+\frac{1}{4}\right|=\frac{1}{6}+\left|2x+\frac{1}{4}\right|\ge\frac{1}{6}\)

\(\Rightarrow\)GTNN  của C là \(\frac{1}{6}\)khi \(\hept{\begin{cases}2x+\frac{1}{4}=0\Leftrightarrow x=\frac{-1}{8}\\\left(x+\frac{1}{2}\right).\left(-\frac{1}{3}-x\right)\ge0\Leftrightarrow\frac{-1}{2}\le x\le\frac{-1}{3}\end{cases}}\)

\(\frac{3x^3+10x^2-9x-4}{\left(x-1\right)\left(x^2+2x-8\right)}=4\)

<=> \(\frac{\left(x+4\right)\left(x-1\right)\left(3x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+4\right)}=4\)

<=> \(\frac{3x+1}{x-2}=4\)

<=> 3x + 1 = 4(x - 2)

<=> 3x + 1 = 4x - 8

<=> 4x - 3x = 8 + 1

<=> x = 9

Vậy tập nghiệm của pt là \(S=\left\{9\right\}\)

(3x - 2)(x² + 5) = (2 - 3x)(5x - 1)

<=> (3x - 2)(x² + 5) + (3x - 2)(5x - 1) = 0

<=> (3x - 2)(x² + 5x + 5 - 1) = 0

<=> (3x - 2)(x² + 5x + 4) = 0

<=> (3x - 2)(x² + x + 4x + 4) = 0

<=> (3x - 2)[x(x + 1) + 4(x + 1)] = 0

<=> (3x - 2)(x + 1)(x + 4) = 0

<=> \(\left\{\begin{matrix}3x-2=0\\x+1=0\\x+4=0\end{matrix}\right.\)

<=> \(\left\{\begin{matrix}x=\frac{2}{3}\\x=-1\\x=-4\end{matrix}\right.\)

Vậy nghiệm nhỏ nhất của pt là - 4

(3x-2)(x²+5)=(2-3x)(5x-1)

=>3x³-2x²+15x-10=10x-15x²-2+3x

=>3x³+13x²+2x-8=0

=>(x-\(\frac{2}{3}\) )(x+4)(x+1)=0

=>\(\left[\begin{matrix}x=\frac{2}{3}\\x=-4\\x=-1\end{matrix}\right.\)

Vậy nghiệm nhỏ nhất của phương trình là x=-4

a/ \(\left|x-\frac{1}{3}\right|=2\frac{1}{5}\)

\(\left|x-\frac{1}{3}\right|=\frac{11}{5}\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{11}{5}\\x-\frac{1}{3}=\frac{-11}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{11}{5}+\frac{1}{3}\\x=\frac{-11}{5}+\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{33}{15}+\frac{5}{15}\\x=\frac{-33}{15}+\frac{5}{15}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{38}{15}\\x=\frac{28}{15}\end{matrix}\right.\)

Vậy:

c/

\(\left|x+\frac{1}{4}\right|-\frac{3}{4}=5\%\)

\(\left|x+\frac{1}{4}\right|-\frac{3}{4}=\frac{1}{20}\)

\(\left|x+\frac{1}{4}\right|=\frac{1}{20}+\frac{15}{20}\)

\(\left|x+\frac{1}{4}\right|=\frac{4}{5}\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{4}{5}\\x+\frac{1}{4}=\frac{-4}{5}\end{matrix}\right.\)

Bạn tự tính tiếp nhé!

sakura Kiemono l là gì vậy

sakura Kiemono \(l\) là gì vậy bạn?

a) \(5x-7=3x+9\)

\(\Rightarrow5x-3x=9+7\)

\(\Rightarrow2x=16\)

\(\Rightarrow x=16:2\)

\(\Rightarrow x=8\)

Vậy \(x=8.\)

b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{2}{5}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{2}{5}-\frac{1}{2}\\x=\left(-\frac{2}{5}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}.\)

c) \(5x-\left|9-7x\right|=3\)

\(\Rightarrow\left|9-7x\right|=5x-3\)

\(\Rightarrow\left[{}\begin{matrix}9-7x=5x-3\\9-7x=3-5x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}9+3=5x+7x\\9-3=-5x+7x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}12=12x\\6=2x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=12:12\\x=6:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{1;3\right\}.\)

d) \(-5+\left|3x-1\right|+6=\left|-4\right|\)

\(\Rightarrow-5+\left|3x-1\right|+6=4\)

\(\Rightarrow-5+\left|3x-1\right|=4-6\)

\(\Rightarrow-5+\left|3x-1\right|=-2\)

\(\Rightarrow\left|3x-1\right|=\left(-2\right)+5\)

\(\Rightarrow\left|3x-1\right|=3.\)

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

Vậy \(x\in\left\{\frac{4}{3};-\frac{2}{3}\right\}.\)

Chúc bạn học tốt!

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