giải phương trình sau: a) \(|x-2|+3x-9=0\)
b) \(\left(x^2-5x+1\right)^2-2x^2+10x=1\)
Giải các phương trình sau:
a) \(|x-2|+3x-9=0\)
b) \(\left(x^2-5x+1\right)^2-2x^2+10x=1\)
a) \(\left|x-2\right|+3x-9=0\)
\(\Leftrightarrow\left|x-2\right|=9-3x\)=> \(9-3x>0\Leftrightarrow x< 3\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=9-3x\\x-2=3x-9\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=11\\-2x=-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{4}\left(tm\right)\\x=\frac{7}{2}\left(ktm\right)\end{cases}}\)
giải các phương trình sau:
a)\(3\left(x^2+x\right)^2-7\left(x^2+x\right)+4=0\)0
b)\(x\left(x+1\right)\left(x^2+x+1\right)=42\)
c)\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
d)\(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)
a) đặt \(\left(x^2+x\right)\)là \(y\)
ta có: \(3y^2-7y+4\)\(=0\)
<=>\(\left(3y-4\right)\left(y-1\right)=0\)
còn lại bạn tự xử nhé
giải các phương trình sau: a) \(|x-2|+3x-9=0\)
b) \(\left(x^2-5x+1\right)^2-2x^2+10x=1\)
mk sắp ik hok rồi, ai giúp vs
a) \(\left|x-2\right|+3x-9=0\)
\(\Leftrightarrow\left|x-2\right|=9-3x\)
+) Xét \(x\ge2\)
\(pt\Leftrightarrow x-2=9-3x\)
\(\Leftrightarrow x+3x=9+2\)
\(\Leftrightarrow4x=11\)
\(\Leftrightarrow x=\frac{11}{4}\left(tm\right)\)
+) Xét \(x< 2\)
\(pt\Leftrightarrow2-x=9-3x\)
\(\Leftrightarrow-x+3x=9-2\)
\(\Leftrightarrow2x=7\)
\(\Leftrightarrow x=\frac{7}{2}\left(ktm\right)\)
Vậy....
Giải các phương trình sau:
\(a.\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(b.\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(c.2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)\)
\(d.\dfrac{7}{8}x-5\left(x-9\right)=\dfrac{20x+1,5}{6}\)
\(e.\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)
\(f.\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
giải phương trình sau:
a)\(2\left(1-x\right)\sqrt{x^2+2x-1}+2x+1=x^2\)
b)\(\sqrt{5x-1}+\sqrt[3]{9-x}=2x^2+3x-1\)
a.
ĐKXĐ: \(x^2+2x-1\ge0\)
\(x^2+2x-1+2\left(x-1\right)\sqrt{x^2+2x-1}-4x=0\)
Đặt \(\sqrt{x^2+2x-1}=t\ge0\)
\(\Rightarrow t^2+2\left(x-1\right)t-4x=0\)
\(\Delta'=\left(x-1\right)^2+4x=\left(x+1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=1-x+x+1=2\\t=1-x-x-1=-2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=2\\\sqrt{x^2+2x-1}=-2x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-5=0\\3x^2-2x+1=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=-1\pm\sqrt{6}\)
b.
ĐKXĐ: \(x\ge\dfrac{1}{5}\)
\(2x^2+x-3+2x-\sqrt{5x-1}+2-\sqrt[3]{9-x}=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3\right)+\dfrac{\left(x-1\right)\left(4x-1\right)}{2x+\sqrt[]{5x-1}}+\dfrac{x-1}{4+2\sqrt[3]{9-x}+\sqrt[3]{\left(9-x\right)^2}}=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3+\dfrac{4x-1}{2x+\sqrt[]{5x-1}}+\dfrac{1}{4+2\sqrt[3]{9-x}+\sqrt[3]{\left(9-x\right)^2}}\right)=0\)
\(\Leftrightarrow x=1\) (ngoặc đằng sau luôn dương)
BÀI 1: GIẢI CÁC PHƯƠNG TRÌNH SAU
a) \(\left(2x-1\right)^2=49\)
b) \(\left(5x-3\right)^2-\left(4x-7\right)^2=0\)
c) \(\left(2x+7\right)^2-9\left(x+3\right)^2\)
d) \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
e) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
f) \(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)
a) 4 b) 0,6 hoặc 1,75 e) -3,5 hoặc -3
Giải các phương trình sau:
a) \(\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)^2=5x^4\)
b) \(2x^4-5x^3-9x^2+11x+4=0\)
c) \(8x^3+4x^2+2x-3=0\)
d) \(\frac{10x^4}{\left(1+x^2\right)^2}-\frac{3x^2}{1+x^2}-1=0\)
e) \(3x^4+4x^3-27x^2+8x+12=0\)
làm tạm câu này vậy
a/\(\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)^2=5x^4\)
\(\Leftrightarrow\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)+4x^4=9x^4\)
\(\Leftrightarrow\left\{\left(x^2-x+1\right)^2+2x^2\right\}=\left(3x^2\right)^2\)
\(\Leftrightarrow\left(x^2-x+1\right)^2+2x^2=3x^2\)(vì 2 vế đều không âm)
\(\Leftrightarrow\left(x^2-x+1\right)=x^2\)
\(\Leftrightarrow\left|x\right|=x^2-x+1\)\(\left(x^2-x+1=\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x=x^2-x+1\\-x=x^2-x+1\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x^2+1=0\left(vo.nghiem\right)\end{cases}}}\)
Vậy...
i cum back <(") câu e/ bạn xét x=0 không là nghiệm của pt, sau đó chia 2 vế cho \(x^2\), đặt ẩn phụ \(t=x+\frac{1}{x}\)rồi giải
Bài tập. Giải các phương trình sau:
a) \(\left|7-x\right|+2x=3\)
b) \(\left|2x-3\right|-4x-9=0\)
c) \(\left|3x+5\right|=\left|2-5x\right|\)
d) \(x\left|x-3\right|-\left|x^2+x+1\right|=1\)
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
Giải các phương trình sau bằng cách đặt ẩn phụ :
a) \(\left(4x-5\right)^2-6\left(4x-5\right)+8=0\)
b) \(\left(x^2+3x-1\right)^2+2\left(x^2+3x-1\right)-8=0\)
c) \(\left(2x^2+x-2\right)^2+10x^2+5x-16=0\)
d) \(\left(x^2-3x+4\right)\left(x^2-3x+2\right)=3\)
e) \(\dfrac{2x^2}{\left(x+1\right)^2}-\dfrac{5x}{x+1}+3=0\)
f) \(x-\sqrt{x-1}-3=0\)