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\(\frac{2}{3}\left(x-1\right)-x-\frac{3}{4}=1\)
<=> \(\frac{2}{3}x-\frac{2}{3}-x-\frac{3}{4}=1\)
<=> \(-\frac{1}{3}x-\frac{17}{12}=1\)
<=> \(-\frac{1}{3}x=\frac{29}{12}\)
<=> \(x=-\frac{29}{4}\)
\(\frac{5}{6}\left(x+2\right)-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(\frac{5}{6}x+\frac{5}{3}-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x+\frac{7}{6}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x=-\frac{5}{6}\)
<=> \(x=5\)
học tốt
(1-4x)^2=9^3
=>(1-4x)^2=729
=>1-8x=729
=>8x=729-1
=>8x=728
=>x=728:8
=>x=91
Vậy x=91
Học Tốt nhé
\(\left(1-4x\right)^2=9^3\)
\(\Rightarrow\left(1-4x\right)^2=729\)
\(\Rightarrow\left(1-4x\right)^2=27^2\)
\(\Rightarrow\orbr{\begin{cases}1-4x=27\\1-4x=-27\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6.5\\x=7\end{cases}}}\)
\(\text{a, 3(x+1)+4x=10}\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow7x+3=10\)
\(\Rightarrow7x=10-3=7\)
\(\Rightarrow x=1\)
c, x+1/10+x+2/9=x+3/8+x+4/7
=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)
=> x+11/10 + x+11/9 = x+11/8 + x+11/7
...............
a) \(3\left(x+1\right)+4x=10\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow3x+4x=10-3\)
\(\Rightarrow7x=7\)
\(\Rightarrow x=7\)
\(b)B=\left|x-\frac{1}{2}\right|+\frac{3}{4}\)
Dùng KT \(\left|x\right|\ge0\)\(\forall\)\(x\)
BG :
Ta có : \(\left|x-\frac{1}{2}\right|\ge0\)\(\forall\)\(x\)
\(\Rightarrow\)\(\left|x-\frac{1}{2}\right|+\frac{3}{4}\ge0+\frac{3}{4}\)\(\forall\)\(x\)
\(\Rightarrow\)\(\left|x-\frac{1}{2}\right|+\frac{3}{4}\ge\frac{3}{4}\)\(\forall\)\(x\)
Hay \(B\ge\frac{3}{4}\)\(\forall\)\(x\)
Dấu "=" xảy ra khi :
\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy GTNN của \(B=\frac{3}{4}\)đạt được khi \(x=\frac{1}{2}\)
\(A=\left|x+\frac{3}{2}\right|\ge0\)
\(MinA=0\Rightarrow\left|x+\frac{3}{2}\right|=0\Rightarrow x=\frac{-3}{2}\)
\(B=\left|x-\frac{1}{2}\right|+\frac{3}{4}\)
\(B\ge\frac{3}{4}\)do\(\left|x-\frac{1}{2}\right|\ge0\)
\(MinB=\frac{3}{4}\Rightarrow\left|x-\frac{1}{2}\right|=0\Rightarrow x=\frac{1}{2}\)
Ta có: \(\left(x-1\right)^2+\left(x-1\right)^3=6\)
\(\Leftrightarrow x^2-2x+1+x^3-3x^2+3x-1-6=0\)
\(\Leftrightarrow x^3-2x^2+x-6=0\)
Thực sự nghiệm PT rất xấu nên bạn xem kỹ lại đề nhé
\(x_1=2,537...\) ; \(x_2=-0,268...\pm1,513...\)
\(\left(x-1\right)^2+\left(x-1\right)^3=6\)
\(\left(x-1\right)^2\left(1+x-1\right)-6=0\)
\(\left(x-1\right)^2\cdot x-6=0\)
\(x\left(x^2-2x+1\right)-6=0\)
\(x^3-2x^2+x-6=0\)
Đến đây bấm máy tính nha