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Câu 1:
a) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
\(\Leftrightarrow\dfrac{12x-2\left(5x+2\right)}{12}=\dfrac{3\left(7-3x\right)}{12}\)
\(\Leftrightarrow12x-10x-4=21-9x\)
\(\Leftrightarrow11x=25\)
\(\Leftrightarrow x=\dfrac{25}{11}\)
b) \(\left(3x-1\right)\left(x-3\right)\left(7-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\Leftrightarrow x=\dfrac{1}{3}\\x-3=0\Leftrightarrow x=3\\7-2x=0\Leftrightarrow x=3,5\end{matrix}\right.\)
c) \(\left|3x\right|=4x+8\) (1)
Ta có: \(\left|3x\right|=3x\Leftrightarrow3x\ge0\Leftrightarrow x\ge0\)
\(\left|3x\right|=-3x\Leftrightarrow3x< 0\Leftrightarrow x< 0\)
Với \(x\ge0\), phương trình (1) có dạng:
\(3x=4x+8\Leftrightarrow-x=8\Leftrightarrow x=-8\)
(không thoả mãn điều kiện) \(\rightarrow\) loại
Với \(x< 0\), phương trình (1) có dạng:
\(-3x=4x+8\Leftrightarrow-7x=8\Leftrightarrow x=-\dfrac{8}{7}\)
(thoả mãn điều kiện) \(\rightarrow\) nhận
Vậy phương trình đã cho có 1 nghiệm \(x=-\dfrac{8}{7}\)
Câu 2:
\(2x\left(6x-1\right)\ge\left(3x-2\right)\left(4x+3\right)\)
\(\Leftrightarrow12x^2-2x\ge12x^2+9x-8x-6\)
\(\Leftrightarrow-3x\ge-6\)
\(\Leftrightarrow x\le2\)
Vậy bất phương trình đã cho có nghiệm \(x\le2\)
a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
a: \(\Leftrightarrow4\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3x^2\)
\(\Leftrightarrow4\cdot\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]=3x^2\)
=>4(x^2+60)^2+132x(x^2+60)+1085x^2=0
=>4(x^2+60)^2+62x(x^2+60)+70x(x^2+60)+1085x^2=0
=>2(x^2+60)(2x^2+120+31x)+35x(2x^2+120+31x)=0
=>(2x^2+120+35x)(2x^2+31x+120)=0
=>\(x\in\left\{\dfrac{-35\pm\sqrt{265}}{4};-\dfrac{15}{2};-8\right\}\)
b: Đặt x^2-3x=a
Phương trình sẽ là \(\dfrac{1}{a+3}+\dfrac{2}{a+4}=\dfrac{6}{a+5}\)
\(\Leftrightarrow\dfrac{a+4+2a+6}{\left(a+3\right)\left(a+4\right)}=\dfrac{6}{a+5}\)
=>(3a+10)(a+5)=6(a^2+7a+12)
=>6a^2+42a+72=3a^2+15a+10a+50
=>3a^2+17a+22=0
=>x=-2 hoặc x=-11/3
Hai câu là hoàn toàn giống nhau, mình làm câu a, câu b bạn tự làm tương tự:
ĐKXĐ: ...
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{4}{4x+\frac{7}{x}-8}+\frac{3}{4x+\frac{7}{x}-10}=1\)
Đặt \(4x+\frac{7}{x}-10=t\)
\(\Leftrightarrow\frac{4}{t+2}+\frac{3}{t}=1\Leftrightarrow4t+3\left(t+2\right)=t\left(t+2\right)\)
\(\Leftrightarrow t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x+\frac{7}{x}-10=-1\\4x+\frac{7}{x}-10=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2-9x+7=0\\4x^2-16x+7=0\end{matrix}\right.\) (bấm casio)
\(\left(\dfrac{4x}{x^2-4x+7}-1\right)+\left(\dfrac{3x}{x^2-5x+7}-1\right)=2\)