\(a,\left(4\frac{1}{2}-\frac{2}{5}x\right):1\frac{3}{4}=\frac{11}{14}\)

\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right):\frac{7}{4}=\frac{11}{4}\)

\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{11}{4}\cdot\frac{7}{4}\)

\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{77}{16}\)

\(\Rightarrow\frac{9}{2}-\frac{2}{5}x=\frac{77}{16}\)

\(\Rightarrow-\frac{2}{5}x=\frac{77}{16}-\frac{9}{2}\)

\(\Rightarrow-\frac{2}{5}x=\frac{5}{16}\)

\(\Rightarrow x=\frac{5}{16}:\left(-\frac{2}{5}\right)\)

\(\Rightarrow x=-\frac{25}{32}\)

\(b,\frac{2}{3}\cdot x-\frac{2}{5}x=\frac{9}{3}\)

\(\Rightarrow x\left(\frac{2}{3}-\frac{2}{5}\right)=\frac{8}{3}\)

\(\Rightarrow x\cdot\frac{4}{15}=\frac{8}{3}\)

\(\Rightarrow x=\frac{8}{3}:\frac{4}{15}\)

\(\Rightarrow x=10\)

\(c,\frac{-2}{3}|x|+1\frac{1}{2}=\frac{2}{5}\)

\(\Rightarrow\frac{-2}{3}|x|+\frac{3}{2}=\frac{2}{5}\)

\(\Rightarrow\frac{-2}{3}|x|=\frac{2}{5}-\frac{3}{2}\)

\(\Rightarrow\frac{-2}{3}|x|=-\frac{11}{10}\)

\(\Rightarrow|x|=\frac{-11}{10}:\frac{-2}{3}\)

\(\Rightarrow|x|=\frac{33}{20}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{33}{20}\\x=-\frac{33}{20}\end{cases}}\)

\(d,|2x-\frac{1}{3}|+\frac{1}{6}=\frac{3}{4}\)

\(\Rightarrow|2x-\frac{1}{3}|=\frac{3}{4}-\frac{1}{6}\)

\(\Rightarrow|2x-\frac{1}{3}|=\frac{7}{12}\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=\frac{7}{12}\\2x-\frac{1}{3}=-\frac{7}{12}\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{11}{12}\\2x=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{24}\\x=-\frac{1}{8}\end{cases}}}\)

Cuối cùng cũng có 2 người trả lời !!!😅😅😅

a,<=>\(\frac{\left(2x+1\right)^2}{4}\)+\(\frac{2\left(2x-1\right)^2}{4}\)≥\(\frac{12\left(x+5\right)^2}{4}\)

<=>4x²+4x+1+2(4x²-4x+1)≥12(x²+10x+25)

<=>4x²+4x+1+8x²-8x+2≥12x²+120x+300

<=>4x²+4x+1+8x²-8x+2-12x²-120x-300≥0

<=>-124x-297≥0

<=>124x+297≤0

<=>124x≤-297

<=>x≤\(\frac{-297}{124}\)

b, Tương tự câu a

c,

TH1: 5-3x=2+x

<=> -3x - x = 2 - 5

<=> -4x = -3

<=> x = 3/4

TH2: 5-3x = -2 - x

<=> -3x + x = -2 - 5

<=> -2x = -7

<=> x = 7/2

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

\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}< \frac{1}{3}\\x-\frac{3}{5}< -\frac{1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{1}{3}+\frac{3}{5}\\x< \frac{-1}{3}+\frac{3}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x< \frac{5}{15}+\frac{9}{15}\\x< \frac{-5}{15}+\frac{9}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)

                vay \(\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)

2) \(\left|x+\frac{11}{2}\right|>\left|-5,5\right|\)

\(\left|x+\frac{11}{2}\right|>5,5\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{11}{2}>\frac{11}{2}\\x+\frac{11}{2}>-\frac{11}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{11}{2}-\frac{11}{2}\\x>\frac{-11}{2}-\frac{11}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)

vay \(\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)

3) \(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)

\(\Rightarrow\left|x-\frac{7}{5}\right|>\frac{2}{5}\) va \(\left|x-\frac{7}{5}\right|< \frac{3}{5}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{7}{5}>\frac{2}{5}\\x-\frac{7}{5}>\frac{-2}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{2}{5}+\frac{7}{5}\\x>\frac{-2}{5}+\frac{7}{5}\end{cases}}\)va \(\orbr{\begin{cases}x-\frac{7}{5}< \frac{3}{5}\\x-\frac{7}{5}< \frac{-3}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{3}{5}+\frac{7}{5}\\x< \frac{-3}{5}+\frac{7}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>\frac{9}{5}\\x>1\end{cases}}\)va \(\orbr{\begin{cases}x< 2\\x< \frac{4}{5}\end{cases}}\)

vay ....

Ahuhu, không ai biết cách giải ư ? T^T

1)

ĐKXĐ: x\(\ne\)3

ta có :

\(\frac{x^2-6x+9}{2x-6}=\frac{\left(x-3\right)^2}{2\left(x-3\right)}=\frac{x-3}{2}\)

để biểu thức A có giá trị = 1

thì :\(\frac{x-3}{2}\)=1

=>x-3 =2

=>x=5(thoả mãn điều kiện xác định)

vậy để biểu thức A có giá trị = 1 thì x=5

1)

\(A=\frac{x^2-6x+9}{2x-6}\)

A xác định

\(\Leftrightarrow2x-6\ne0\)

\(\Leftrightarrow2x\ne6\)

\(\Leftrightarrow x\ne3\)

Để A = 1

\(\Leftrightarrow x^2-6x+9=2x-6\)

\(\Leftrightarrow x^2-6x-2x=-6-9\)

\(\Leftrightarrow x^2-8x=-15\)

\(\Leftrightarrow x=3\) (loại vì không thỏa mãn ĐKXĐ)

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 :

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

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

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

\(\Leftrightarrow15x=3\)

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

<=> \(\frac{2x-3}{35}\)+ \(\frac{x^2-2x}{7}\)< \(\frac{5x^2}{35}\)- \(\frac{7\left(2x-5\right)}{35}\)

<=> \(\frac{2x-3}{35}\)+ \(\frac{5\left(x^2-2x\right)}{35}\)< \(\frac{5x^2}{35}\)- \(\frac{7\left(2x-5\right)}{35}\)

<=> 2x - 3 + 5( x² - 2x ) < 5x² - 7( 2x - 5 )

<=> 2x - 3 + 5x² - 10x < 5x² - 14x + 35

<=> 2x + 5x² - 10x - 5x² + 14x < 35 + 3

<=> 6x < 38

<=> x < \(\frac{19}{3}\)

có sai thì thông cảm :)